Cubes And Cube Roots

(i) To calculate the cube of (8)³

We have to multiply the given number three times;

= (8 × 8 × 8) = 512

So, 512 is the cube of 8.

(ii) (15)³

First multiply the given number three times;

= (15 × 15 ×15) = 3375

So, 3375 is the cube of 15

(iii) (21)³

First multiply the given number three times;

= (21 × 21 × 21) = 9261

9261 is the cube of 21.

(iv) (60)³

First multiply the given number three times;

= (60 × 60 × 60) = 216000

216000 is the cube of 60.

(i) To calculate the cube of (1.2)³

We have to multiply the given number three times;

= (1.2 × 1.2 × 1.2) = 1.728

Now by converting it into fraction we get,

=

(ii) To calculate the cube of (3.5)³

We have to multiply the given three times;

= (3.5 × 3.5 × 3.5) = 42.875

Now by converting it into fraction we get,

=

(iii) To calculate the cube of (0.8)³

We have to multiply the given number by its power;

= (0.8 × 0.8 × 0.8) = 0.512

Now by converting it into fraction we get,

=

(iv) To calculate the cube of (0.05)³

We have to multiply the given number by its power;

= (0.05 × 0.05 × 0.05) = 0.000125

Now by converting it into fraction we get,

=

(i)

By multiplying we get,

So, cube of

(ii)

Multiplying the given number three times we get,

(iii)

Multiplying the given number three times we get,

(iv)

Multiplying the given number three times we get,

(i) 125

First find out the prime factors of 125,

125 = 5×5×5

As we see a group of three 5 is made, which we can also be write as 5³;

So, 125 is the product of triplets of 5.

Therefore, it is the perfect cube.

(ii) 243

The prime factorization of 256 is shown below:

243 = 3 × 3 × 3 × 3 × 3

To be a perfect cube the prime factors of number should make a group of 3 but as we can see here more than 3 numbers are available in prime factors.

So, 243 is not the perfect cube.

(iii) 343

The prime factorization of 256 is shown below:

343 = 7 × 7 × 7

As we see a group of three 7 is formed, which we can also be write as 7³;

So, 343 is the product of triplets of 7.

Therefore, it is the perfect cube.

(iv) 256

The prime factorization of 256 is shown below:

256 = 2×2×2×2×2×2×2×2

If the prime factors are not making the pairs of three so the number is not perfect cube.

(v) 8000

The prime factorization of 8000 is shown below:

8000 = 2×2×2×2×2×2×5×5×5

As we can see three pairs can be made of the above prime factors, which are 2³, 2³, and 5³.

So, 8000 can be expressed as the product of the triplets of 2, 2 and 5, i.e.

2³ × 2³ × 5³ = 20³

Therefore, 8000 is a perfect cube.

(vi) 9261

The prime factorization of 9261 is shown below:

9261 = 3×3×3×7×7×7

As we can see two pairs can be made of the above prime factors, which are 3³, and 7³.

So, 9261 can be expressed as the product of the triplets of 3 and 7, i.e.

3³ × 7³ = 21³

Therefore, 9261 is a perfect cube.

(vii) 5324

The prime factorization of 5324 is shown below:

5324 = 2×2×11×11×11

Therefore, 5324 is not a perfect cube.

(viii) 3375

The prime factorization of 3375 is shown below:

3375 = 3×3×3×5×5×5

As we can see two pairs can be made of the above prime factors, which are 3³, and 5³.

So, 3375 can be expressed as the product of the triplets of 3 and 5, i.e.

3³ × 5³ = 15³

Therefore, 3375 is a perfect cube.

By the rule for even numbers, the cubes of even numbers are always even.

So, first we have to look for which given numbers are even.

216, 512 and 1000 are the even numbers.

Now, the prime factorization are as follows:

∴ 216 = 2×2×2×3×3×3 = 2³ × 3³ = 6³

∴ 512= 2×2×2×2×2×2×2×2×2 = 2³ × 2³ × 2³ = 8³

∴ 1000 = 2×2×2×5×5×5 = 2³ × 5³ = 10³

Therefore, we can say that 216, 512 and 1000 are the cube of even numbers.

By the rule for odd numbers, the cubes of odd numbers are always odd.

So, first we have to look for which of the given numbers are odd.

125, 343 and 9261 are the odd numbers.

∴ 125 = 5×5×5 = 5³

∴ 343 = 7×7×7 = 7³

∴ 9261 = 3×3×3×7×7×7 = 3³ × 7³ = 21³

As we can see, odd numbers are the cubes of odd number.

Therefore, 125, 343 and 9261 are the cubes of odd numbers.

Let’s find out the prime factors of the given number,

1323 = 3 × 3 × 3 × 7 × 7

As we can see, one 7 is required to make the pair of two triplets. So, 7 will be the smallest number to multiply 1323 to make it the perfect cube.

Let’s find out the prime factors of the given number,

∴ 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

As we can see, to make the pair of 4 triplets two 5 are required, which is 5×5.

So, 25 will be the number multiplied to 2560, to get the perfect cube.

Let’s find out the prime factors of the given number,

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

So, in these two pairs of 2 triplets, two 5 are extra. Therefore to get the perfect cube we have to divide the given number by 5×5, which is 25.

Let’s find out the prime factors of the given number,

8788 = 2 × 2 × 13 × 13 × 13

So, in this pair of triplets, two 2 are extra. Therefore to get the perfect cube we have to divide the given number by 2×2, which is 4.

Let’s take, a = 2 and b = 5

So, by using the formula,

We have a³+3a²b+3ab²+b³

Keep the digits

By taking the highlighted digits only,

We get;

؞ (25)³ = 15625

Let’s take, a = 4 and b = 7

So, by using the formula,

We have, a³ + 3a²b + 3ab² + b³

By taking the highlighted digits only,

We get;

؞(47)³ = 103823

Let’s take,

a = 6 and

b = 8

So, by using the formula,

We have a³+3a²b+3ab²+b³

By taking the highlighted digits only,

We get;

؞(68)³ = 314432

Let’s take,

a = 8 and

b = 4

So, by using the formula,

We have a³+3a²b+3ab²+b³

By taking the highlighted digits only,

We get;

؞(84)³ = 592704

By prime factorization method we get;

64 = 2×2×2×2×2×2

= (2×2×2) × (2×2×2)

؞ = (2×2) = 4

By prime factorization method we get;

343 = 7×7×7

= (7×7×7)

= 7

By prime factorization method we get;

729 = 3×3×3×3×3×3

= (3×3×3) × (3×3×3)

= (3×3) = 9

By prime factorization method we get;

1728 = 2×2×2×2×2×2×3×3×3

= (2×2×2) × (2×2×2) × (3×3×3)

= 2³ × 2³ × 3³

= (2×2×3) = 12

By prime factorization method we get;

9261 = 3×3×3×7×7×7

= (3×3×3) × (7×7×7)

= (3×7) = 21

By prime factorization method we get;

4096 = 2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2 ×2

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

= 2³ × 2³ × 2³ × 2³

= (2 ×2 ×2 ×2) = 16

By prime factorization method we get;

8000 = 2 ×2 ×2 ×2 ×2 ×2 ×5 ×5 ×5

= (2 ×2 ×2) × (2 ×2 ×2) × (5 ×5 ×5)

= 2³ × 2³ × 5³

= (2 ×2 ×5) = 20

By prime factorization method we get;

3375 = 3 ×3 ×3 ×5 ×5 ×5

= (3 ×3 ×3) × (5 ×5 ×5)

= (3 ×5) = 15

By prime factorization method we get;

216 = 2 ×2 ×2 ×3 ×3 ×3

= (2 ×2 ×2) × (3 ×3 ×3)

= - (2 ×3) = - 6

By prime factorization method we get;

512 = 2×2×2×2×2×2×2×2×2

= (2×2×2) × (2×2×2) × (2×2×2)

= - 8

By prime factorization method we get;

1331 = 11×11×11 = 11³

= -11

By prime factorization method we get;

27;

27 = 3×3×3

64;

64 = 2×2×2×2×2×2

Now,

By prime factorization method we get;

125

216

By prime factorization method we get;

27 = 3×3×3

125

By prime factorization method we get;

64 = 2 ×2 ×2 ×2 ×2 ×2

343;

By prime factorization method we get;

64 = 2×2×2×2×2×2

= (4) × (9) = 36

By prime factorization method we get;

The prime factorization of 512:

The prime factorization of 343:

The prime factorization of 141:

=

The prime factorization of 294:

=

The prime factorization of 216:

=

The prime factorization of 496:

=

So, only in case of 216 a perfect pair of three 6’s is formed, Hence 216 is a perfect cube.

Having prime factors of all the numbers we get,

=

4 (not a perfect cube)

=

(perfect cube)

=

(not a perfect cube)

= no prime factors are possible

Hence, only case of 1331 a perfect triplet pair of 11’s is formed, Hence 1331 is a perfect cube.

The prime factorization of 512 is :

The prime factorization of 125 is :

125 = 5×5×5

The prime factorization of 64 is :

64 = 2×2×2×2×2×2

=

64 = 2 × 2 × 2 × 2 × 2 × 2

343 = 7 × 7 × 7

=

The prime factorization of 512 is :

The prime factorization of 729 is :

=

The prime factorization of 648 is :

648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2 × 2 × 2 × 3 × 3 × 3 × 3

So, we can see that to make it a perfect cube we should complete the pair of three 3’s .

Hence, it will be multiplied by = 3 × 3 = 9.

The prime factorization of 1536 is :

1536 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

So, we can see that to make it a perfect cube we should eliminate one 3 from it.

Hence, it will be divided by = 3.

We have:

We have:

0.8 × 0.8 × 0.8 = 0.512.

The prime factorization of 4096 is:

=

The prime factorization of 216 is :

The prime factorization of 343 is:

343 = 7×7×7

=

64 = 2 × 2 × 2 × 2 × 2 × 2

125,

=

=

.

(i) =

(ii) =

(iii) = .

(iv) = 0.5³

= 0.5 × 0.5 × 0.5

= 0.125.

From the prime factorization of 121, we get,

121 = 11 × 11

169 = 13 × 13

196 = 2 × 2 × 7 × 7 = 14 × 14

216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 3 = 6

Hence, in case of 216 perfect triplet pairs of 2’s and 3’s is formed, so 216 is a perfect cube.

216 = 2 × 2 × 2 × 3 × 3 × 3

64 = 2 × 2 × 2 × 2 × 2 × 2

=

343 = 7 × 7 × 7

=

=

From the prime factorization of 324 we get,

= 324 = 2 × 2 × 3 × 3 × 3 × 3 = 3 × 3 × 3 × 2 × 2 × 3

So to make it a perfect cube we must complete the triplet pairs of 2’s and 3’s.

Hence, it will be multiplied by 2 × 3 × 3 = 18.

From the prime factorization of 128, we get,

From the prime factorization of 250, we get,

=

From the prime factorization of 216, we get,

216 = 2×2×2×3×3×3

From the prime factorization of 512, we get,

From the prime factorization of 343, we get,

343 = 7×7×7

From the prime factorization of 1000, we get,

216 = 6 × 6 × 6

512 = 8 × 8 × 8

343 = 7 × 7 × 7

1000 = 10 × 10 × 10

Hence, we can see clearly that 343 is a cube of an odd number.