Cubes

Let us understand, what perfect cube is.

A perfect cube is a number that is a cube of an integer.

So,

To find the perfect cube from the following numbers, we need to find their prime factors individually and then group them into triplets of the prime factors.

If no prime factor is left out, then the number is a perfect cube.

However, if one of the prime factors is a single factor or double factor, then the number is not a perfect cube.

Take 125.

Prime factors of 125 = 5 × 5 × 5

Group them into triplets.

⇒ Prime factors of 125 = (5 × 5 × 5)

Since, prime factor 5 form a triplet, so 125 is a perfect cube.

∴, 125 is a perfect cube.

Take 500.

Prime factors of 500 = 5 × 5 × 5 × 2 × 2

Group the factors into triplets.

⇒ Prime factors of 500 = (5 × 5 × 5) × 2 × 2

Since, prime factor 2 is not part of a triplet, so 500 is not a perfect cube.

∴, 500 is not a perfect cube.

Take 64.

Prime factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

Group the factors into triplets.

⇒ Prime factors of 64 = (2 × 2 × 2) × (2 × 2 × 2)

Since, prime factor 2 forms two triplets, so 64 is a perfect cube.

∴, 64 is a perfect cube.

Take 7³.

Prime factors of 7³ = 7 × 7 × 7

Group the factors into triplets.

⇒ Prime factors of 7³ = (7 × 7 × 7)

Since, prime factor 7 forms a triplets, so 7³ is a perfect cube.

∴, 7³ is a perfect cube.

Take 729.

Prime factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

Group the factors into triplets.

⇒ Prime factors of 729 = (3 × 3 × 3) × (3 × 3 × 3)

Since, prime factor 3 forms two triplets, so 729 is a perfect cube.

∴, 729 is a perfect cube.

Take 968.

Prime factors of 968 = 2 × 2 × 2 × 11 × 11

Group the factors into triplets.

⇒ Prime factors of 968 = (2 × 2 × 2) × 11 × 11

Since, prime factor 11 is not a part of a triplet, so 968 is not a perfect cube.

∴, 968 is not a perfect cube.

We have to form two cubes of lengths 5 cm and 1 cm.

We know, a cube is symmetrical three-dimensional shape, either solid or hollow, contained by six equal squares.

We need to find the number of small cubes to form the big cube.

For this, we need to find the volume of the big cube.

Volume is the amount of space that a substance or object occupies, or that is enclosed within a container, and that is why we need to find volume to further find number of small cubes in that big cube.

Take the cube of length 5 cm.

Volume of cube of length 5 cm is given as

Volume = length × length × length

Since, length = 5 cm

⇒ Volume = 5 × 5 × 5 cm³

⇒ Volume = 125 cm³

This means, in the cube of length 5 cm, we have 125 cm cubes.

Now, take the cube of length 1 cm.

Volume of cube of length 1 cm is given as

Volume = length × length × length

Since, length = 1 cm

⇒ Volume = 1 × 1 × 1 cm³

⇒ Volume = 1 cm³

This means, in the cube of length 1 cm, we have 1 cm cubes.

Given that,

Sumanta has made many cubes of length, 1 cm.

This means, volume of each cube = length × length × length

Here, length = 1 cm

⇒ Volume of each cube = 1 × 1 × 1 cm³

⇒ Volume of each cube = 1 cm³

And a single cube out of those many cubes is made by Manami.

Let us assume that Sumanta has made ‘Z’ cubes of length 1 cm, where Z is some natural number.

So, if volume of 1 cube = 1 cm³

⇒ Volume of Z cubes = Z × 1 cm³

⇒ Volume of Z cubes = Z cm³

We have volumes of cube in the options.

(i). Let us check for 100.

We need to find cube root of 100. If we are able to find the cube root of 100, then we can say that a big cube is possible in this case.

100 = 2×2× 5× 5

We can see 100 cannot be written as a perfect cube.

∴ 100 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 100 is not a natural number.

Thus, Manami won’t be able to make a big cube in this case.

(ii). Let us check for 1000.

We need to find the cube root of 1000. If we are able to find the cube root of 1000, then we can say that a big cube is possible in this case.

Cube root of 1000

1000 = 10× 10× 10

⇒ Cube root of 1000 = (1000)^1/3

= 10

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1000 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(iii). Let us check for 1331.

We need to find the cube root of 1331. If we are able to find the cube root of 1331, then we can say that a big cube is possible in this case.

Cube root of 1331

1331 = 11× 11× 11

⇒ Cube root of 1331 = (1331)^1/3

= 11

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1331 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(iv). Let us check for 1210.

We need to find the cube root of 1210. If we are able to find the cube root of 1210, then we can say that a big cube is possible in this case.

Cube root of 1210

1210 = 2× 5× 11× 11

We can see 1210 cannot be written as a perfect cube.

∴ 1210 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 1210 is not a natural number.

Thus, Manami will not be able to make a big cube in this case.

(v). Let us check for 3375.

We need to find the cube root of 3375. If we are able to find the cube root of 3375, then we can say that a big cube is possible in this case.

Cube root of 3375

3375 = 15× 15× 15

⇒ Cube root of 3375 = (3375)^1/3

= 15

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 3375 is a natural number.

Thus, Manami will be able to make a big cube in this case.

(vi). Let us check for 2700.

We need to find the cube root of 2700. If we are able to find the cube root of 2700, we can say that a big cube is possible in this case.

Cube root of 2700

2700 = 3× 3× 3× 2× 2× 5× 5

We can see 2700 cannot be written as a perfect cube.

∴ 2700 ≠ (any integer)³

Since, we have cubes of length 1 cm each.

So, length of the final big cube will also be a natural number.

And cube root of 2700 is not a natural number.

Thus, Manami will not be able to make a big cube in this case.

Let us understand what perfect cube is.

A perfect cube is a number that is the cube of an integer.

(i). Let us check for 216.

We need to check whether 216 is a cube of an integer or not.

Let us factorize 216 for ease of calculation.

We have,

216 = 2 × 2 × 2 × 3 × 3 × 3

Group these factors into three similar integers.

216 = (2 × 2 × 2) × (3 × 3 × 3)

Take cube root on both sides,

Cube root of 216 is an integer.

Thus, 216 is a perfect cube.

(ii). Let us check for 343.

We need to find whether 343 is cube of an integer or not.

Let us factorize 343 for ease of calculation.

We have,

343 = 7 × 7 × 7

Group these factors into three similar integers.

343 = (7 × 7 × 7)

Take cube root on both sides,

Cube root of 343 is an integer.

Thus, 343 is a perfect cube.

(iii). Let us check for 1024.

We need to find whether 1024 is cube of an integer or not.

Let us factorize 1024 for ease of calculation.

We have,

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Group these factors into three similar integers.

1024 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × 2

Take cube root on both sides,

Cube root of 1024 is clearly not an integer, since is not an integer.

Thus, 1024 is not a perfect cube.

(iv). Let us check for 324.

We need to find whether 324 is cube of an integer or not.

Let us factorize 324 for ease of calculation.

We have,

324 = 2 × 2 × 3 × 3 × 3 × 3

Group these factors into three similar integers.

324 = (3 × 3 × 3) × 2 × 2 × 3

Take cube root on both sides,

Cube root of 324 is clearly not an integer, since is not an integer.

Thus, 324 is not a perfect cube.

(v). Let us check for 1744.

We need to find whether 1744 is cube of an integer or not.

Let us factorize 1744 for ease of calculation.

We have,

1744 = 2 × 2 × 2 × 2 × 109

Group these factors into three similar integers.

1744 = (2 × 2 × 2) × 2 × 109

Take cube root on both sides,

Cube root of 1744 is clearly not an integer, since is not an integer.

Thus, 1744 is not a perfect cube.

(vi). Let us check for 1372.

We need to find whether 1372 is cube of an integer or not.

Let us factorize 1372 for ease of calculation.

We have,

1372 = 2 × 2 × 7 × 7 × 7

Group these factors into three similar integers.

1372 = (7 × 7 × 7) × 2 × 2

Take cube root on both sides,

Cube root of 1372 is clearly not an integer, since is not an integer.

Thus, 1372 is not a perfect cube.

Given that,

Debnath made a cuboid of dimension:

Length = 4 cm

Breadth = 3 cm

Height = 2 cm

Since, volume of cuboid = length × breadth × height

⇒ Volume of cuboid = 4 × 3 × 2

= 24

Therefore, volume of a cuboid made by Debnath is 24 cm³.

Factors of 24 are:

24 = 2 × 2 × 2 × 3

This 24 cm³ is volume of one cuboid, which is not even perfect cube.

And to form a cube, we need to have perfect cube volumes.

So make 24 into a perfect cube.

Notice the factors of 24,

24 = 2 × 2 × 2 × 3

Group three similar integers, we get

24 = (2 × 2 × 2) × 3

Multiply both sides by (3 × 3), we get

24 × (3 × 3) = (2 × 2 × 2) × 3 × (3 × 3)

⇒ 216 = (2 × 2 × 2) × (3 × 3 × 3)

Number of cuboids = (3 × 3)

Because multiplying (3 × 3) makes a cube by the same number of cuboids.

⇒ Number of cuboids = 9

Thus, number of required cuboids are 9.

Let us recall what a perfect cube is.

A perfect cube is a number that is the cube of an integer.

(i). For 675.

We need to factorize 675. We get,

675 = 3 × 3 × 3 × 5 × 5

Group these factors into three similar integers. We have

675 = (3 × 3 × 3) × 5 × 5

To make it into a perfect cube, multiply both sides by 5.

We have

675 × 5 = (3 × 3 × 3) × 5 × 5 × 5

Now, group these factors into three similar integers.

3375 = (3 × 3 × 3) × (5 × 5 × 5)

We are able to group it in three similar integers. This means, 3375 is a perfect cube.

Thus, smallest positive number that should be multiplied with 675 is 5.

(ii). For 200.

We need to factorize 200. We get,

200 = 2 × 2 × 2 × 5 × 5

Group these factors into three similar integers. We have

200 = (2 × 2 × 2) × 5 × 5

To make it into perfect cube, multiply both sides by 5.

We have

200 × 5 = (2 × 2 × 2) × 5 × 5 × 5

Now, group these factors into three similar integers.

1000 = (2 × 2 × 2) × (5 × 5 × 5)

We are able to group it in three similar integers. This means, 1000 is a perfect cube.

Thus, smallest positive number that should be multiplied with 200 is 5.

(iii). For 108.

We need to factorize 108. We get,

108 = 2 × 2 × 3 × 3 × 3

Group these factors into three similar integers. We have

108 = (3 × 3 × 3) × 2 × 2

To make it into perfect cube, multiply both sides by 2.

We have

108 × 2 = (3 × 3 × 3) × 2 × 2 × 2

Now, group these factors into three similar integers.

216 = (3 × 3 × 3) × (2 × 2 × 2)

We are able to group it in three similar integers. This means, 216 is a perfect cube.

Thus, smallest positive number should be multiplied with 108 is 2.

(iv). For 121.

We need to factorize 121. We get,

121 = 11 × 11

We can’t group it into three similar integers. So, multiply both sides by 11.

121 × 11 = 11 × 11 × 11

Now, group these factors into three similar integers.

1331 = (11 × 11 × 11)

Since, we are able to group it in three similar integers. This means, 1331 is a perfect cube.

Thus, smallest positive number that should be multiplied with 121 is 11.

(v). For 1225.

We need to factorize 1225. We get,

1225 = 5 × 5 × 7 × 7

We can’t group it into three similar integers. So, multiply both sides by (5 × 7).

1225 × 5 × 7 = 5 × 5 × 7 × 7 × (5 × 7)

Now, group these factors into three similar integers.

42875 = (5 × 5 × 5) × (7 × 7 × 7)

Since, we are able to group it in three similar integers. This means, 42875 is a perfect cube.

Thus, smallest positive number that should be multiplied with 42875 is (5 × 7), i.e., 35.

Let us recall what a perfect cube is.

A perfect cube is a number that is the cube of an integer.

(i). For 7000.

We need to factorize 7000. We get,

7000 = 7 × 2 × 5 × 2 × 5 × 2 × 5

Group these factors into three similar integers.

7000 = (2 × 2 × 2) × (5 × 5 × 5) × 7

To make it into perfect cube, divide it by 7 on both sides.

⇒ 1000 = (2 × 2 × 2) × (5 × 5 × 5)

Since, we are able to group it in three similar integers. This means, 1000 is a perfect cube.

Thus, smallest positive number that should be divided with 7000 is 7.

(ii). For 2662.

We need to factorize 2662. We get,

2662 = 2 × 11 × 11 × 11

Group these factors into three similar integers.

2662 = (11 × 11 × 11) × 2

To make it into perfect cube, divide it by 2 on both sides.

⇒ 1331 = (11 × 11 × 11)

Since, we are able to group it in three similar integers. This means, 1331 is a perfect cube.

Thus, smallest positive number that should be divided with 2662 is 2.

(iii). For 4394.

We need to factorize 4394. We get,

4394 = 2 × 13 × 13 × 13

Group these factors into three similar integers.

4394 = (13 × 13 × 13) × 2

To make it into perfect cube, divide it by 2 on both sides.

⇒ 2197 = (13 × 13 × 13)

Since, we are able to group it in three similar integers. This means, 2197 is a perfect cube.

Thus, smallest positive number that should be divided with 4394 is 2.

(iv). For 6750.

We need to factorize 6750. We get,

6750 = 5 × 5 × 5 × 3 × 3 × 3 × 2

Group these factors into three similar integers.

6750 = (5 × 5 × 5) × (3 × 3 × 3) × 2

To make it into perfect cube, divide it by 2 on both sides.

⇒ 3375 = (5 × 5 × 5) × (3 × 3 × 3)

Since, we are able to group it in three similar integers. This means, 3375 is a perfect cube.

Thus, smallest positive number that should be divided with 6750 is 2.

(v). For 675.

We need to factorize 675. We get,

675 = 5 × 5 × 3 × 3 × 3

Group these factors into three similar integers.

675 = (3 × 3 × 3) × 5 × 5

To make it into perfect cube, divide it by (5 × 5) on both sides.

⇒ 27 = (3 × 3 × 3)

Since, we are able to group it in three similar integers. This means, 27 is a perfect cube.

Thus, smallest positive number that should be divided with 675 is (5 × 5), i.e., 25.

Let us recall the definition of prime factors and cube roots.

Prime factors are any of the prime numbers that can be multiplied to give the original number.

The cube root of a number is a special value that, when used in a multiplication three times, gives that number.

(i). For 512:

512 can be divided by prime number 2, quotient is 256.

256 can be divided by prime number 2, quotient is 128.

128 can be divided by prime number 2, quotient is 64.

64 can be divided by prime number 2, quotient is 32.

32 can be divided by prime number 2, quotient is 16.

16 can be divided by prime number 2, quotient is 8.

8 can be divided by prime number 2, quotient is 4.

4 can be divided by prime number 2, quotient is 2.

2 can be divided by prime number 2, quotient is 1.

So, 512 can be expressed in product of prime factors as:

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Group these factors into three similar integers.

512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

Taking cube root on both sides, we get

Thus, cube root of 512 is 8.

(ii). For 1728.

1728 can be divided by prime number 2, quotient is 864.

864 can be divided by prime number 2, quotient is 432.

432 can be divided by prime number 2, quotient is 216.

216 can be divided by prime number 2, quotient is 108.

108 can be divided by prime number 2, quotient is 54.

54 can be divided by prime number 2, quotient is 27.

27 can be divided by prime number 3, quotient is 9.

9 can be divided by prime number 3, quotient is 3.

3 can be divided by prime number 3, quotient is 1.

So, 1728 can be expressed in product of prime factors as:

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

Group these factors into three similar integers.

1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)

Taking cube root on both sides, we get

Thus, cube root of 1728 is 12.

(iii). For 5832.

5832 can be divided by prime number 2, quotient is 2916.

2916 can be divided by prime number 2, quotient is 1458.

1458 can be divided by prime number 2, quotient is 729.

729 can be divided by prime number 3, quotient is 243.

243 can be divided by prime number 3, quotient is 81.

81 can be divided by prime number 3, quotient is 27.

27 can be divided by prime number 3, quotient is 9.

9 can be divided by prime number 3, quotient is 3.

3 can be divided by prime number 3, quotient is 1.

So, 5832 can be expressed in product of prime factors as:

5832 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Group these factors into three similar integers.

5832 = (2 × 2 × 2) × (3 × 3 × 3) × (3 × 3 × 3)

Taking cube root on both sides, we get

Thus, cube root of 5832 is 18.

(iv). For 15625.

15625 can be divided by prime number 5, quotient is 3125.

3125 can be divided by prime number 5, quotient is 625.

625 can be divided by prime number 5, quotient is 125.

125 can be divided by prime number 5, quotient is 25.

25 can be divided by prime number 5, quotient is 5.

5 can be divided by prime number 5, quotient is 1.

So, 15625 can be expressed in product of prime factors as:

15625 = 5 × 5 × 5 × 5 × 5 × 5

Group these factors into three similar integers.

15625 = (5 × 5 × 5) × (5 × 5 × 5)

Taking cube root on both sides, we get

Thus, cube root of 15625 is 25.

(v). For 10648.

10648 can be divided by prime number 2, quotient is 5324.

5324 can be divided by prime number 2, quotient is 2662.

2662 can be divided by prime number 2, quotient is 1331.

1331 can be divided by prime number 11, quotient is 121.

121 can be divided by prime number 11, quotient is 11.

11 can be divided by prime number 11, quotient is 1.

So, 10648 can be expressed in product of prime factors as:

10648 = 2 × 2 × 2 × 11 × 11 × 11

Group these factors into three similar integers.

10648 = (2 × 2 × 2) × (11 × 11 × 11)

Taking cube root on both sides, we get

Thus, cube root of 10648 is 22.

Given is, lengths of a side of cubes.

We need to find volumes of these cubes.

Let us recall the formula of volume of cube.

Volume of cube = length × length × length

⇒ Volume of cube = (length)³

(i). Given is,

Length of cube (unit) = p² + q²

Then,

Volume of cube = (p² + q²)³

Recall the algebraic identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

Put a = p² and b = q². We get

(p² + q²)³ = (p²)³ + (q²)³ + 3(p²)²(q²) + 3(p²) (q²)²

⇒ (p² + q²)³ = p⁶ + q⁶ + 3p⁴q² + 3p²q⁴

⇒ (p² + q²)³= p⁶ + q⁶ + 3p²q² (p² + q²)

= p⁶ + q⁶ + 3p⁴q² + 3p²q⁴

Thus, volume of cube is p⁶ + q⁶ + 3p⁴q² + 3p²q⁴cubic unit.

(ii). Given is,

Then,

Recall the algebraic identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

Put and . We get

Thus, volume of cube is cubic unit.

(iii). Given is,

Length of cube (unit) = x²y – z²

Then,

Volume of cube = (x²y – z²)³

Recall the algebraic identity,

(a – b)³ = a³ – b³ – 3a²b + 3ab²

Put a = x²y and b = z². We get

(x²y – z²)³ = (x²y)³ – (z²)³ – 3(x²y)²(z²) + 3(x²y)(z²)²

= x⁶y³ – z⁶ – 3x⁴y²z² + 3x²yz⁴

= x⁶y³ – z⁶ – 3x²yz² (x²y – z²)

= x⁶y³ – z⁶ – 3x⁴y²z² + 3x² y z⁴

Thus, volume of cube is x⁶y³ – z⁶ – 3x⁴y²z² + 3x² y z⁴ cubic unit.

(iv). Given is,

Length of cube (unit) = 1 + b – 2c

Then,

Volume of cube = (1 + b – 2c)³

Recall the algebraic identity,

(a + b + c)³ = a³ + b³ + c³ + 3(a + b) (b + c)(c + a)

Put a = 1, b = b and c = -2c. We get

(1 + b – 2c)³ = (1)³ + (b)³ + (-2c)³ + 3(1 + b) (b – 2c)(-2c + 1)

⇒ (1 + b – 2c)³ = 1 + b³ – 8c³ + 3(1 + b) (b – 2c)(1 – 2c)

Thus, volume of cube is [1 + b³ – 8c³ + 3(1 + b)(b – 2c)(1 – 2c)] cubic unit.

(v). Given is,

Volume of cube (cubic unit) = (2.89)³ + (2.11)³ + 15 × 2.89 × 2.11

We need to find length of cube.

Volume of cube can be written as,

Volume = (2.89)³ + (2.11)³ + [3 × 5 × 2.89 × 2.11]

[∵, 15 = 3 × 5]

⇒ Volume = (2.89)³ + (2.11)³ + [3 × (2.89 + 2.11) × 2.89 × 2.11]

[∵, 5 = 2.89 + 2.11]

Put 2.89 = a and 2.11 = b. We get

Volume = a³ + b³ + [3 × (a + b) × a × b]

Or, Volume = a³ + b³ + 3ab (a + b)

Or, Volume = (a + b)³ …(a)

[∵, by algebraic identity, we know that

(a + b)³ = a³ + b³ + 3ab (a + b)]

Again,

Put a = 2.89 and b = 2.11. We get

Volume = (2.89 + 2.11)³

⇒ Volume = (5)³

We know, volume of cube is given by

Volume = (length)³

Or (length)³ = Volume

⇒ length = (Volume)^1/3

⇒ length = 5

Thus, length of cube is 5 unit.

(vi). Given is,

Volume of cube (cubic unit) = (2m + 3n)³ + (2m – 3n)³ + 12m (4m² – 9n²)

We need to find length of cube.

Volume of cube can be written as,

[∵, By identity, (a + b) (a – b) = a² – b² in (2m + 3n) (2m – 3n)]

Volume = (2m + 3n)³ + (2m – 3n)³ + [3 × 4m × (2m + 3n) (2m – 3n)]

[∵, 4m = 2m + 2m]

= (2m + 3n)³ + (2m – 3n)³ + [3 × (2m + 2m) × (2m + 3n) (2m – 3n)]

[∵, 3n – 3n = 0 and it won’t affect the equation]

⇒ Volume = (2m + 3n)³ + (2m – 3n)³ + [3 × (2m + 2m + 3n – 3n) × (2m + 3n) (2m – 3n)]

[∵, (2m + 2m + 3n – 3n) = (2m + 3n) + (2m – 3n); we have just rearranged]

⇒ Volume = (2m + 3n)³ + (2m – 3n)³ + [3 × ((2m + 3n) + (2m – 3n)) × (2m + 3n)(2m – 3n)]
⇒ Volume = (2m + 3n)³ + (2m – 3n)³ + 3[(2m + 3n) + (2m – 3n)][(2m + 3n)(2m – 3n)]

Put (2m + 3n) = a and (2m – 3n) = b. We get

Volume = a³ + b³ + 3(a + b) ab

Or, Volume = a³ + b³ + 3ab (a + b)

Or, Volume = (a + b)³ …(a)

[∵, by algebraic identity, we know that

(a + b)³ = a³ + b³ + 3ab (a + b)]

Again,

Put a = (2m + 3n) and b = (2m – 3n) in equation (a). We get

Volume = ((2m + 3n) + (2m – 3n))³

⇒ Volume = (2m + 3n + 2m – 3n)³

⇒ Volume = (4m)³

We know, volume of cube is given by

Volume = (length)³

Or (length)³ = Volume

⇒ length = (Volume)^1/3

⇒ length = 4m

Thus, length of cube is 4m unit.

(vii). Given is,

Volume of cube (cubic unit) = (a + b)³ – (a – b)³ – 6b(a² – b²)

We need to find the length of cube.

Volume of cube can be written as,

Volume = (a + b)³ – (a – b)³ – [3 × 2b × (a² – b²)]

[∵, 6b = 3 × 2b]

⇒ Volume = (a + b)³ – (a – b)³ – [3 × 2b × (a + b)(a – b)]

[∵, By identity, (a² – b²) = (a + b)(a – b)]

⇒ Volume = (a + b)³ – (a – b)³ – [3 × (b + b) × (a + b)(a – b)]

[∵, 2b = b + b]

⇒ Volume = (a + b)³ – (a – b)³ – [3 × (b + b + a – a) × (a + b)(a – b)]

[∵, a – a = 0 and it won’t affect the equation]

⇒ Volume = (a + b)³ – (a – b)³ – [3 × ((a + b) + (-a + b)) × (a + b)(a – b)]

[∵, (b + b + a – a) = ((a + b) + (-a + b)); we have just rearranged]

⇒ Volume = (a + b)³ – (a – b)³ – [3 × ((a + b) – (a – b)) × (a + b)(a – b)]

[∵, (-a + b) = -(a – b)]

⇒ Volume = (a + b)³ – (a – b)³ – 3((a + b) – (a – b))(a + b)(a – b)

Put (a + b) = x and (a – b) = y. We get

Volume = x³ – y³ – 3(x – y)xy

Or, Volume = x³ – y³ – 3xy(x – y)

Or, Volume = (x – y)³ …(A)

[∵, by algebraic identity, we know that

(x – y)³ = x³ – y³ – 3xy(x – y)]

Again,

Put x = (a + b) and y = (a – b) in equation (A). We get

Volume = ((a + b) – (a – b))³

⇒ Volume = (a + b – a + b)³

⇒ Volume = (2b)³

We know, volume of cube is given by

Volume = (length)³

Or (length)³ = Volume

⇒ length = (Volume)^1/3

⇒ length = 2b

Thus, length of cube is 2b unit.

(viii). Given is,

Length of cube = 2x – 3y – 4z

Then,

Volume of cube = (2x – 3y – 4z)³

Recall the algebraic identity,

(a + b + c)³ = a³ + b³ + c³ + 3(a + b + c)(ab + bc + ca) – 3abc

Put a = 2x, b = -3y and c = -4z. We get

(2x – 3y – 4z)³ = (2x)³ + (-3y)³ + (-4z)³ + 3(2x – 3y – 4z)((2x)(-3y) + (-3y)(-4z) + (-4z)(2x)) – 3(2x)(-3y)(-4z)

⇒ (2x – 3y – 4z)³ = 8x³ – 27y³ – 64z³ + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz

Thus, volume of cube is [8x³ – 27y³ – 64z³ + 3(2x – 3y – 4z)(-6xy + 12yz – 8zx) – 72xyz] cubic unit.

(ix). Given is,

Volume of cube (cubic unit) = x⁶ – 15x⁴ + 75x² – 125

We need to find length of cube.

Volume of cube can be written as,

[∵, x⁶ = (x²)³, 15 = 3 × 5, 75 = 3 × 25 and 125 = (5)³]

Volume = (x²)³ – (3 × 5x⁴) + (3 × 25x²) – (5)³

⇒ Volume = (x²)³ – (3 × 5 × (x²)²) + (3 × (5)² × x²) – (5)³

[∵, x⁴ = (x²)² and 25 = (5)²]

⇒ Volume = (x²)³ + (3 × -5 × (x²)²) + (3 × (-5)² × x²) + (-5)³ …(A)

[∵, (5)² = (-5)² and -(5)³ = (-5)³]

Put x² = a and (-5) = b in equation (A). We get

Volume = a³ + (3 × b × a²) + (3 × b² × a) + b³

Or, Volume = a³ + 3ba² + 3b²a + b³

Or, Volume = (a + b)³

[∵, by algebraic identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²]

Put a = x² and b = (-5). We get

Volume = (x² – 5)³

We know, volume of cube is given by

Volume = (length)³

Or (length)³ = Volume

⇒ length = (Volume)^1/3

⇒ length = x² – 5

Thus, length of cube is (x² – 5) unit.

(x). Given is,

Volume of cube (cubic unit) = 1000 + 30x(10 + x) + x³

We need to find length of cube.

Volume of cube can be written as,

Volume = (10)³ + 30x(10 + x) + x³

[∵, 1000 = (10)³]

⇒ Volume = (10)³ + [3 × 10x × (10 + x)] + x³ …(A)

[∵, 30 = 3 × 10]

Put 10 = a and x = b in equation (A). We get

Volume = a³ + [3 × ab × (a + b)] + b³

Or, Volume = a³ + b³ + 3ab(a + b)

Or, Volume = (a + b)³

[∵, by algebraic identity,

(a + b)³ = a³ + b³ + 3ab(a + b)]

Put a = 10 and b = x. We get

Volume = (10 + x)³

We know, volume of cube is given by

Volume = (length)³

Or (length)³ = Volume

⇒ length = (Volume)^1/3

⇒ length = 10 + x

Thus, length of cube is (10 + x) unit.

Let us solve.

(a). Given that, x – y = 2.

We need to find the value of x³ – y³ – 6xy.

By algebraic identity,

(a – b)³ = a³ – b³ – 3a²b + 3ab²

Or, (a – b)³ = a³ – b³ – 3ab(a – b)

Or, a³ – b³ = (a – b)³ + 3ab(a – b)

Put a = x and b = y. We get

x³ – y³ = (x – y)³ + 3xy(x – y) …(i)

Take x³ – y³ – 6xy.

x³ – y³ – 6xy = (x³ – y³) – 6xy

⇒ x³ – y³ – 6xy = ((x – y)³ + 3xy(x – y)) – 6xy [from (i)]

⇒ x³ – y³ – 6xy = ((2)³ + 3xy(2)) – 6xy [∵, Given that, (x – y) = 2]

⇒ x³ – y³ – 6xy = (8 + 6xy) – 6xy [∵, 2³ = 8]

⇒ x³ – y³ – 6xy = 8 + 6xy – 6xy

⇒ x³ – y³ – 6xy = 8 + 0 [∵, 6xy – 6xy = 0]

⇒ x³ – y³ – 6xy = 8

Thus, x³ – y³ – 6xy = 8.

(b). Given:

To prove:

Proof: We know the algebraic identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

Or, (a + b)³ = a³ + b³ + 3ab(a + b)

Or, a³ + b³ = (a + b)³ – 3ab(a + b) …(i)

Take a³ + b³ – ab.

a³ + b³ – ab = (a³ + b³) – ab

⇒ a³ + b³ – ab = ((a + b)³ – 3ab(a + b)) – ab

[∵, we know that, ]

Thus, .

Hence proved.

(c). Given that, x + y = 2 and

Let us solve this further.

We have,

⇒ x + y = 2xy

⇒ 2xy = x + y

⇒ 2xy = 2 [∵, we know that, (x + y) = 2]

⇒ xy = 1

Now, we have

x + y = 2 …(i)

xy = 1 …(ii)

We need to find the value of x³ + y³.

By algebraic identity,

(x + y)³ = x³ + y³ + 3x²y + 3xy²

Or, (x + y)³ = x³ + y³ + 3xy(x + y)

Or, x³ + y³ = (x + y)³ – 3xy(x + y)

Substituting value of (x + y) and xy from equation (i) and (ii) respectively, we get

x³ + y³ = (2)³ – 3(1)(2)

⇒ x³ + y³ = 8 – 6

⇒ x³ + y³ = 2

Thus, x³ + y³ = 2.

(d). Given that,

…(i)

We need to find the value of

Further simplifying (i), we get

x² – 1 = 2x

Now, take cube on both sides. We get

(x² – 1)³ = (2x)³

[∵, we have the identity,

(a – b)³ = a³ – b³ – 3a²b + 3ab²]

⇒ (x²)³ – 1³ – 3(x²)² + 3x² = (2x)³

⇒ x⁶ – 1 – 3x⁴ + 3x² = 8x³

Dividing both sides by x³,

Thus, .

(e). Given that,

We need to find the value of .

Let us take .

Take cube on both sides of this equation. We get

[∵, from the identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²]

Thus, .

(f). Given that,

x = y + z

We need to find the value of x³ – y³ – z³ – 3xyz.

Take x = y + z.

We can write as,

x – y – z = 0 …(i)

Now, take cube on both sides of the above equation. We get,

(x – y – z)³ = 0

Since, by the identity

(x – y – z)³ = x³ – y³ – z³ + 3(x – y – z)(-xy + yz – zx) – 3xyz

Rearranging it,

x³ – y³ – z³ – 3xyz = (x – y – z)³ – 3(x – y – z)(-xy + yz – zx)

⇒ x³ – y³ – z³ – 3xyz = (0)³ – 3(0)(-xy + yz – zx) [from equation (i)]

⇒ x³ – y³ – z³ – 3xyz = 0 – 0

⇒ x³ – y³ – z³ – 3xyz = 0

Thus, x³ – y³ – z³ – 3xyz = 0.

(g). Given: xy(x + y) = m

To prove:

Proof: Take xy(x + y) = m.

Rearranging it,

Taking cube on both sides, we get

By identity, we have

(x + y)³ = x³ + y³ + 3x²y + 3xy²

So,

[∵, given that, xy(x + y) = m]

Hence, proved.

(h). Given:

To prove:

Proof: We have,

Putting and .

⇒ p + q = 4pq …(i)

We have,

Multiplying p and q, we get

…(ii)

Now, take .

[∵, (p + q)³ = p³ + q³ + 3p²q + 3pq²

Or, p³ + q³ = (p + q)³ – 3p²q – 3pq²

Or, p³ + q³ = (p + q)³ – 3pq(p + q)]

…(iii)

Substituting the value of pq from equation (ii) in equation (iii), we get

Hence, proved.

(i). Given that,

We need to find the value of .

We have,

…(i)

Taking cube on both sides of the equation, we get

Now, adding 2 on both sides of this equation, we get

Thus, .

(j). Given that,

a³ + b³ + c³ = 3abc

where a ≠ b ≠ c.

We need to find the value of a + b + c.

We have,

a³ + b³ + c³ = 3abc

⇒ a³ + b³ + c³ – 3abc = 0

We know that,

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

⇒ 0 = (a + b + c)(a² + b² + c² – ab – bc – ca)

⇒ (a + b + c)(a² + b² + c² – ab – bc – ca) = 0

This means,

Either (a + b + c) = 0

Or (a² + b² + c² – ab – bc – ca) = 0

If a² + b² + c² – ab – bc – ca = 0

Multiplying by 2 on both sides, we get

2(a² + b² + c² – ab – bc – ca) = 2 × 0

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

⇒ a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0

⇒ (a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² + 2ca) = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0

[∵, by identity, we know (x – y)² = x² + y² – 2xy]

This means,

(a – b)² = 0 ⇒ a = b

(b – c)² = 0 ⇒ b = c

(c – a)² = 0 ⇒ c = a

But, a ≠ b ≠ c.

⇒ (a² + b² + c² – ab – bc – ca) ≠ 0

This clearly means,

a + b + c = 0

Thus, a + b + c = 0.

(k). Given that,

m + n = 5 …(i)

mn = 6 …(ii)

We need to find the value of (m² + n²)(m³ + n³).

By identity, we know

(m + n)³ = m³ + n³ + 3mn(m + n)

Or, m³ + n³ = (m + n)³ – 3mn(m + n) …(iii)

Also, by identity,

(m + n)² = m² + n² + 2mn

Or, m² + n² = (m + n)² – 2mn …(iv)

Take (m² + n²)(m³ + n³).

So,

(m² + n²)(m³ + n³) = ((m + n)² – 2mn)((m + n)³ – 3mn(m + n))

Substituting values of (m + n) and mn from equation (i) and (ii) respectively in the above equation, we get

(m² + n²)(m³ + n³) = ((5)² – 2(6))((5)³ – 3(6)(5))

⇒ (m² + n²)(m³ + n³) = (25 – 12)(125 – 90)

⇒ (m² + n²)(m³ + n³) = 13 × 35

⇒ (m² + n²)(m³ + n³) = 455

Thus, (m² + n²)(m³ + n³) = 455.

(i)

Expression 1: x + 9

Expression 2: x² – 9x + 81

Product

= (x + 9)(x² – 9x + 81)

= (x + 9)(x² – 9x + 9²)

= x³ + 9³ [∵ (a + b)(a² – ab + b²) = a³ + b³]

= x³ + 729

(ii)

Expression 1: 2a – 1

Product = 8a³ – 1

= (2a)³ - 1³

= (2a – 1)[(2a)² + (2a)(1) + (1)²]

[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (2a – 1)(4a² + 2a + 1)

Hence, Expression 2 is (4a² + 2a + 1)

(iii)

Expression 1: 3 – 5c

Product = 27 – 125c³

= (3)³ – (5c)³

= (3 – 5c)(3² + 3(5c) + (5c)²) [∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (3 – 5c)(9 + 15c + 25c²)

Hence, Expression 2 is (9 + 15c + 25c²)

(iv)

Expression 1: (a + b + c)

Expression 2: (a + b)² – (a + b)c + c²

Product

= (a + b + c)[(a + b)² – (a + b)c + c²]

= [(a + b) + c][(a + b)² – (a + b)c + c²]

[Using, a³ + b³ = (a + b)(a² – ab + b²)]

If a = a + b, b = c

= (a + b)³ + c³

(v)

Expression 1: 3x

Expression 2: (2x – 1)² – (2x – 1) (x + 1) + (x + 1)²

Product

= 3x [(2x – 1)² – (2x – 1)(x + 1) + (x + 1)²]

As 3x can be written as 2x-1+x+1

= [2x-1+x+1] [(2x – 1)² – (2x – 1) (x + 1) + (x + 1)²]

[Using, a³ + b³ = (a + b) (a² – ab + b²)

Here a = 2x-1, b = x+1

= (2x – 1)³ + (x + 1)³

As (a-b)³ =a³ -3a²b+3ab² – b³

(a + b)³=a³+3a²b+3ab²+b³

= (2x)³ – 1 – 3(2x)²(1) + 3(2x)(1)² + x³ + 1 + 3x² + 3x

= 8x³ – 1 – 12x² + 6x + x³ + 1 + 3x² + 3x

= 9x³ – 9x² + 9x

(vi)

Expression 1:

Expression 2:

Product

[using, (a + b)³ = (a + b)(a² – ab + b²)]

(vii)

Expression 1: 4a – 5b

Expression 2: 16a² + 20ab + 25b²

Product

= (4a – 5b)(16a² + 20ab + 25b²)

= (4a – 5b)((4a)² + (4a)(5b) + (5b)²)

[Using, a³ – b³ = (a – b)(a² + ab + b²)]

= (4a)³ – (5b)³

= 64a³ – 125b³

(viii)

Expression 2: a²b² + abcd + c²d²

Product

= a³b³ – c³d³

= (ab)³ – (cd)³

= (ab – cd)[(ab)² + (ab)(cd) + (cd)²]

[Using, a³ – b³ = (a – b)(a² + ab + b²)]

= (ab – cd)(a²b² + abcd + c²d²]

Hence, Expression 1 is (ab – cd)

(ix)

Expression 1: 1 – 4y

Product = 1 – 64y³

= 1³ – (4y)³

= (1 – 4y)(1² + 1(4y) + (4y)²)

[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (1 – 4y)(1 + 4y + 16y²)

Hence, Expression 2 is (1 + 4y + 16y²)

(x)

Expression 1: (2p + 1)

Product = 8(p – 3)³ + 343

= (2(p – 3))³ + 7³

[Using, a³ – b³ = (a – b)(a² + ab + b²)]

= [2(p – 3) + 7][(2(p – 3))² + 2(p – 3)(7) + 7²]

= (2p – 6 + 7)[4(p – 3)² + 14p – 42 + 49]

= (2p + 1)(4(p² – 6p + 9) + 14p + 7)

= (2p + 1)(4p² – 24p + 36 + 14p + 7)

= (2p + 1)(4p² – 10p + 43)

Hence, Expression 2 is (4p² – 10p + 43)

(xi)

Expression 1 : m – p

Expression 2: (m + n)² + (m + n)(n + p) + (n + p)²

Product

= (m – p)[(m + n)² + (m + n)(n + p) + (n + p)²]

= [m + n – (n + p)]((m + n)² + (m + n)(n + p) + (n + p)²]

[Using, a³ – b³ = (a – b)(a² + ab + b²)]

= (m + n)³ – (n + p)³

= (m³ + n³ + 3m²n + 3mn²) – (n³ + p³ + 3n²p + 3np²)

= m³ + n³ + 3m²n + 3mn² – n³ – p³ – 3n²p – 3np²

= m³ – p³ + 3m²n + 3mn² – 3n²p – 3np²

(xii)

Expression 1: (3a - 2b)² + (3a – 2b) × (2a – 3b) + (2a – 3b)²

Expression 2: (a + b)

Product

= (a + b)[(3a - 2b)² + (3a – 2b) × (2a – 3b) + (2a – 3b)²]

= (3a – 2b – (2a – 3b))[(3a - 2b)² + (3a – 2b) × (2a – 3b) + (2a – 3b)²]

= (3a – 2b)³ - (2a – 3b)³

[Now,

(a – b)³ = a³ – b³ – 3a²b + 3ab²]

= [(3a)³ – (2b)³ – 3(3a)²(2b) + 3(3a)(2b)²] – [(2a)³ – (3b)³ – 3(2a)²(3b) + 3(2a)(3b)²]

= [27a³ – 8b³ – 54a²b + 36ab²] – [8a³ – 27b³ – 36a²b + 54ab²]

= 27a³ – 8b³ – 54a²b + 36ab² – 8a³ + 27b³ + 36a²b – 54ab²

= 19a³ + 19b³ – 18a²b – 18b²

_{i.(a + b)(a – b)(a}² + ab + b²)(a² – ab + b²)

= (a + b)(a² – ab + b²)(a – b)(a² + ab + b²)

Using x³ – y³ = (x – y)(x² + xy + y²)

x³ + y³ = (x + y)(x² – xy + y²)

= (a³ + b³)(a³ – b³)

Now, using (a + b)(a – b) = a² – b²

= a⁶ – b⁶

ii. (a – 2b)(a² + 2ab + 4b²)(a³ + 8b³)

= (a – 2b)(a² + a(2b) + (2b)²)(a³ + (2b)³)

Using x³ – y³ = (x – y)(x² + xy + y²)

x³ + y³ = (x + y)(x² – xy + y²)

= (a³ – (2b)³)(a³ + (2b)³)

Now, using (a + b)(a – b) = a² – b²

= a⁶ – (2b)⁶

= a⁶ – 64b⁶

iii. (4a² – 9)(4a² – 6a + 9)(4a² + 6a + 9)

= [(2a)² – 3²](4a² – 6a + 9)(4a² + 6a + 9)

Using x² – y² = (x – y)(x + y)

= (2a – 3)(2a + 3)(4a² – 6a + 9)(4a² + 6a + 9)

= (2a – 3)(4a² + 6a + 9)(2a + 3)(4a² – 6a + 9)

Using x³ – y³ = (x – y)(x² + xy + y²)

x³ + y³ = (x + y)(x² – xy + y²)

= [(2a)³ – 3³][(2a)³ + 3³]

= (2a)⁶ - 3⁶

= 64a⁶ – 729

iv. (x – y)(x² + xy + y²) + (y – z)(y² + yz + z²) + (z – x)(z² + zx + x²)

Using a³ – b³ = (a – b)(a² + ab + b²)

= x³ – y³ + y³ – z³ + z³ – x³

= 0

v. (x + 1)(x² - x + 1) + (2x – 1)(4x² + 2x + 1) – (x – 1)(x² + x + 1)

= (x + 1)(x² - x + 1) + (2x – 1)((2x)² + 2x + 1) – (x – 1)(x² + x + 1)

Using x³ – y³ = (x – y)(x² + xy + y²)

x³ + y³ = (x + y)(x² – xy + y²)

= x³ + 1 + (2x)³ – 1 - (x³ – 1)

= x³ + 1 + 8x³ – 1 – x³ + 1

= 8x³ + 1

Given,

Multiplying both side by ‘x’

⇒ x² + 1 = -x

⇒ x² + x + 1 = 0

Now,

x³ – 1

Using x³ – y³ = (x – y)(x² + xy + y²)

= (x – 1)(x² + x + 1)

= (x – 1) × 0

= 0

Given,

Multiplying both side by ‘a’

⇒ a² + 9 = 3a

⇒ a² – 3a + 9 = 0

Now,

a³ + 27

= a³ + 3³

Using x³ + y³ = (x + y) (x² – xy + y²)

= (a + 3) (a² – 3a + 3²)

= (a + 3) (a² – 3a + 9)

= (a + 3) × 0

= 0

Given,

Multiplying both side by ‘ab’

⇒ a² + b² = ab

⇒ a² – ab + b² = 0

Now,

a³ + b³

= (a + b)(a² – ab + b²)

= (a + b) × 0

= 0

(i) 1000a³+27b⁶

= (10a)³ + (3b²)⁶

Using x³ + y³ = (x = y)(x² - xy + y²)

= (10a + 3b²)[(10a)² - (10a)(3b²) + (3b²)²]

= (10a + 3b²)(100a² - 30ab² + 9b⁴)

(ii) 1-216 z³

= 1 – (6z)³

Using x³ – y³ = (x – y)(x² + xy + y²)

= (1 – 6z)(1² + 1(6z) + (6z)²)

= (1 – 6z)(1 + 6z + 36z²)

(iii) m⁴ - m

= m(m³ – 1)

= m(m³ – 1³)

Using x³ – y³ = (x – y)(x² + xy + y²)

= m(m – 1)(m² + m + 1)

(iv) 192a³+3

= 3(64a³ + 1)

= 3[(4a)³ + 1³]

Using x³ + y³ = (x + y)(x² – xy + y²)

= 3(4a + 1)[(4a)² – 4a + 1]

= 3(4a + 1)(16a² – 4a + 1)

(v) 16a⁴x³ + 54ay³

= 2a(8a³x³ + 27y³)

= 2a[(2ax)³ + (3y)³]

Using x³ + y³ = (x + y)(x² – xy + y²)

= 2a(2ax + 3y)[(2ax)² – 2ax(3y) + (3y)²]

= 2a(2ax + 3y)(4a²x² – 6axy + 9y²)

(vi) 729a³ b³ c³-125

= (9abc)³ - 5³

= (9abc – 5)[(9abc)² + 9abc(5) + 5²]

= (9abc – 5)(81a²b²c² + 45abc + 25)

(vii)

Using x³ - y³ = (x - y)(x² + xy + y²)

(viii)

Using x³ – y³ = (x – y)(x² + xy + y²)

(ix)x³ + 3x²y + 3xy² + 2y³

= x³ + y³ + 3x²y + 3xy² + y³

= (x + y)³ + y³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

= (x + y + y)[(x + y)² - (x + y)y + y²]

[Using x³ – y³ = (x + y)(x² - xy + y²)]

= (x + 2y)(x² + y² - xy - y² + y²)

= (x + 2y)(x² – xy + y²)

(x) 1 + 9x + 27x² + 28x³

= 27x³ + 1 + 27x² + 9x + x³

= (3x)³ + 1³ + 3(3x)³(1) + 3(3x)(1)²

= (3x + 1)³ + x³

[∵ a³ + b³ + 3a²b + 3ab² = (a + b)³]

Now, Using x³ + y³ = (x + y)(x² - xy + y²)

= (3x + 1 + x)[(3x + 1)² - (3x + 1)x + x²]

= (4x + 1)(9x² + 6x + 1 - 3x² - x + x²)

= (4x + 1)(7x² + 5x + 1)

(xi) x³ – 9y³ – 3xy(x-y)

x³ – y³ – 3xy (x – y) – 8y³

[∵ x³ – y³ = x³ – y³ – 3xy (x – y)]

= (x – y)³ – (2y)³

= (x – y – (2y)) [(x – y)² + (x – y) (2y) + (2y)²]

[Using x³ + y³ = (x + y) (x² – xy + y²)]

= (x -3 y) (x² + y² – 2xy + 2xy – 2y² + 4y²)

= (x - 3y) (x² + 3y²)

(xii)8 – a³ + 3a²b – 3ab² + b³

= b³ – a³ – 3ab² + 3a²b + 8

Now, As (x + y)³ = x³ + y³ + 3x²y + 3xy²

= (b – a)³ + 2³

Using x³ + y³ = (x + y) (x² - xy + y²)

= (b – a + 2) [(b – a)² - (b – a)2 + 2²]

= (b – a + 2) (b² + a² – 2ab - 2b + 2a + 4)

(xiii) x⁶+3x⁴ b²+3x² b⁴+b⁶+a³b³

_{= (x}²)³ + (b²)³ + 3(x²)²(b²) + 3(x²)(b²)² + a³b³

_{As (x + y)}³ = x³ + y³ + 3x²y + 3xy²

_{= (x}² + b²)³ + (ab)³

_{Using x}³ + y³ = (x + y)(x² – xy + y²)

_{= (x}² + b² + ab)[(x² + b²)² - (x² + b²)ab + a²b²]

(xiv) x⁶ + 27

= (x²)³ + (3)³

Using x³ + y³ = (x + y)(x² – xy + y²)

= (x² + 3)[(x²)² - 3x² + 3²)

= (x² + 3)(x⁴ - 3x² + 9)

(xv)x⁶ – y⁶

= (x³)² – (y³)²

Using (a² – b²) = (a – b)(a + b)

= (x³ – y³)(x³ + y³)

Using x³ – y³ = (x – y)(x² + xy + y²) and

x³ + y³ = (x + y)(x² – xy + y²)

= (x – y)(x² + xy + y²)(x + y)(x² – xy + y²)

(xvi) x¹² – y¹²

= (x⁶)² – (y⁶)²

Using (a² – b²) = (a – b)(a + b)

= (x⁶ – y⁶)(x⁶ + y⁶)

= [(x³)² – (y³)^{2 �}][(x²)³ + (y²)³]

Using (a² – b²) = (a – b)(a + b)

= (x³ – y³)(x³ + y³)[(x²)³ + (y²)³]

Using x³ – y³ = (x – y)(x² + xy + y²) and

x³ + y³ = (x + y)(x² – xy + y²)

= (x – y)(x² + xy + y²)(x + y)(x² – xy + y²)(x² + y²)(x⁴ - x²y² + x⁶y⁶)

(xvii) m³ -n³-m(m²-n² )+n(m-n)²

Using x³ – y³ = (x – y)(x² + xy + y²) and

Using (a² – b²) = (a – b)(a + b)

= (m – n)(m² + mn + n²) – m(m – n)(m + n) + n(m – n)²

Taking (m – n) as common, we get

= (m – n)[m² + mn + n² – m(m + n) + n(m – n)]

= (m – n)(m² + mn + n² – m² – mn + mn – n²)

= (m – n)mn