Algebra

(5x + 2y + 3z)²

The identity is

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

∴ (5x + 2y + 3z)²

= (5x)² + (2y)² + (3z)² + 2(5x)(2y) + 2(2y)(3z) + 2(3z)(5x)

= 25x² + 4y² + 9z² + 20xy + 12yz + 30zx

(2a + 3b – c)²

Here the identity used is

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(2a + 3b – c)² = (2a)² + (3b)² + (– c)² + 2(2a)(3b) + 2(3b)(– c) + 2(– c)(2a)

= 4a² + 9b² + c² + 12ab – 6bc – 4ca

(x – 2y – 4z)²

Here the identity used is

(x + y + z) ² = x² + y² + z² + 2xy + 2yz + 2zx

∴ (x – 2y – 4z) ²

= x² + (– 2y)² + (– 4z)² + 2(x)(– 2y) + 2(– 2y)(– 4z) + 2 (– 4z)(x)

= x² + 4y² + 16z² – 4xy + 16yz – 8yz

(P – 2q + r)²

Here we use the identity of

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

∴ (p – 2q + r)²

= (p)² + (– 2q)² + (r)² + 2(p)(– 2q) + 2(– 2q)(r) + 2(r)(p)

= p² + 4q² + r² – pq – 4qr + 2rp

(x + 1) (x + 4) (x + 7)

we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x + 1)(x + 4)(x + 7)

= x³ + (1 + 4 + 7) x² + (1 × 4 + 4 × 7 + 7 × 1) x + 1 × 4 × 7

= x³ + 12x² + (4 + 28 + 7) x + 28

= x³ + 12x² + 39x + 28

(p + 2)(p – 4)(p + 6)

Here the identity used is

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (p + 2)(p – 4)(p + 6)

= p³ + (2 – 4 + 6) p² + (2(– 4) + (– 4)6 + 6(2))p + (2)(– 4)(6)

= p³ + 4p² + (– 8 – 24 + 12)p – 48

= p³ + 4p² – 20p – 48

(x + 5)(x – 3)(x – 1)

Here the identity used is

(x + a)(x + b)(x + c) = x³ + (a + b + c) x² + (ab + bc + ca)x + abc

∴ (x + 5)(x – 3)(x – 1)

= x³ + (5 – 3 – 1) x² + (5(– 3) + (– 3)(– 1)(– 1)5)x + 5(– 3)(– 1)

= x³ + x² + (– 15 + 3 – 5) x + 15

= x³ + x² – 17x + 15

(x – a)(x – 2a)(x – 4a)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x – a)(x – 2a)(x – 4a)

= x³ + (– a – 2a – 3a)x² + [(– a)(– 2a) + (– 2a)(– 3a) + (– 3a)(– a)]x + (– a)(– 2a)(– 3a)

= x³ – 6ax² + (2a² + 6a² + 3a²)x – 6a³

= x³ – 6ax² + 11a²x – 6a³

(3x + 1)(3x + 2)((3x + 5)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (3x + 1)(3x + 2)((3x + 5)

= (3x)³ + (1 + 2 + 5)(3x)² + (1 × 2 + 2 × 5 + 5 × 1)(3x) + 1 × 2 × 5

= 27 x³ + (8)9x² + (2 + 10 + 5) (3x) + 1 × 2 × 5

= 27 x³ + 72x² + 51x + 10

(2x + 3)(2x – 5)(2x – 7)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (2x + 3)(2x – 5)(2x – 7)

= (2x)³ + (3 – 5 – 7)(2x)² + [(3)(– 5) + (– 5)(– 7) + (– 7)(3)](2x) + (3)(– 5)(– 7)

= 8x³ + (– 9)(4x²) + (– 15 + 35 – 21)(2x) + 105

= 8x³ – 36 x² – 2x + 105

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x + 7)(x + 3)(x + 9)

= x³ + (7 + 3 + 9)x² + (7 × 3 + 3 × 9 + 9 × 7)x + 7 × 3 × 9

= x³ + 19x² + (21 + 27 + 63)x + 189

= x³ + 19x² + 111x + 189

Coefficient of

x² = 19

x = 111

Constant term = 189

(x – 5)(x – 4)(x + 2)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x – 5)(x – 4)(x + 2)

= x³ + (– 5 – 4 + 2)x² + {(– 5) × (– 4) + (– 4) × 2 + 2 × (– 5)}x + (– 5)(– 4)2

= x³ – 7x² + (20 – 8 – 10)x + 40

= x³ – 7x² + 2x + 40

Coefficient of

x² = – 7

x = 2 and

Constant term = 40

(2x + 3)(2x + 5)(2x + 7)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)(x)² + (ab + bc + ca)x + abc

∴ (2x + 3) (2x + 5)(2x + 7)

= (2x)³ + 4(3 + 5 + 7)x² + (15 + 35 + 21)2x + 105

= 8x³ + 60x² + 142x + 105

Coefficient of

x² = 60

x = 142

Constant term = 105

(5x + 2)(1 – 5x)(5x + 3)

Here we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ – (5x + 2)(5x – 1)(5x + 3)

here x = 5x, a = 2, b = – 1 and c = 3

= – (5x)³ – (2 – 1 + 3)(5x)² – (– 2 – 3 + 6)5x + 6

= – 125x³ – 100x² – 5x + 6

Here coefficient of

x² = – 100

x = – 5 and

constant term = 6

Here, we have to use the identity

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

Comparing this with

x³ – 10x² + 45x – 15 = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

⇒ – 10 = a + b + c

Now,

And (ab + bc + ca) = 45

Now (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ (– 10)² – 2 × 45 = a² + b² + c²

⇒ 100 – 90 = a² + b² + c²

⇒ a² + b² + c² = 10

(i) (3a + 5b)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(3a + 5b)³

= (3a)³ + 3 (3a)²(5b) + 3(3a)(5b)² + (5b)³

= 27a³ + 135a²b + 225ab² + 125b³

(ii) (4x – 3y)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(4x – 3y)³

= (4x)³ + 3(4x)²(– 3y) + 3 × 4x (– 3y)² + (– 3y)³

⁼ 64x³ – 192 x² + 108y² – 27y³

(iii)

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

=

=

(i) (100 – 1)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(100 – 1)³

= (100)³ + 3(100)² (– 1) + 3(100) (– 1)² + (– 1)³

= 1000000 – 30000 + 300 – 1

= 970299

(ii) 101³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(101)³ = (100 + 1)³

= 100³ + 3 × 100² × 1 + 3 × 100 × 1² + 1³

= 1000000 + 30000 + 300 + 1

= 1030301

(iii) 98³

= (100 – 2)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(100 – 2)³

= 100³ + (– 2)³ + 3(100)²(– 2) + 3(100)(– 2)²

= 1000000 – 8 – 60000 + 1200

= 941192

(iv) (102)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(100 + 2)³

= 100³ + 3(100)² (2) + 3(2)²(100) + (2)³

= 1000000 + 60000 + 1200 + 8

= 1061208

(vi) (1002)³

= (1000 + 2)³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(1000 + 2)³

= 1000³ + 3 × 1000²(2) + 3 × 1000 × (2)² + 2³

= 1000000000 + 6000000 + 12000 + 8

= 1006012008

Given 2x + 3y = 13

⇒ (2x + 3y)³ = 13³

Here the identity used is

(a + b)³ = a³ + 3a²b + 3ab² + b³

(2x + 3y)³ = (2x)³ + 3(2x)²3y + 3(2x)(3y)² + (3y)³

⇒ 8x³ + 36x²y + 54xy² + 27y³ = 2197

⇒ 8x³ + 27y³ + 18xy(2x + 3y) = 2197

⇒ 8x³ + 27y³ = 2197 – 108 (2x + 3y) (∵ xy = 6 given)

⇒ 8x³ + 27y³ = 2197 – 108 × 13

⇒ 8x³ + 27y³ = 2197 – 1404

⇒ 8x³ + 27y³ = 793

The identity used here is

(a + b)³ = a³ + 3a²b + 3ab² + b³

∴ (x – y)³ = x³ + 3x²(– y) + 3x(– y)² + (– y)³

⇒ x³ – 3x²y + 3xy² – y³

⇒ (x – y)³ = x³ – 3x²y + 3xy² – y³

⇒ (– 6)³ = x³ – y³ + 3xy(– x + y)

⇒ – 216 + 3 × 4 (– 6) = x³ – y³

⇒ – 216 – 72 = x³ – y³

⇒ – 288 = x³ – y³

Here we use the identity of

(a + b)³ = a³ + 3a²b + 3ab² + b³

∴

⇒

⇒

⇒

Here the identity used

(a + b)³ = a³ + 3a²b + 3ab² + b³

∴

⇒

⇒

⇒

(i) (2x + y + 4z) (4x² + y² + 16z² – 2xy – 4yz – 8zx)

⇒ Here the identity used is

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

Comparing this with given condition

(2x + y + 4z) (4x² + y² + 16z² – 2xy – 4yz – 8zx) = (2x)³ + y³ + (4z)³ – 3 × 2x × y × 4z

⇒ 8x³ + y³ + 64z³ – 24xyz

(ii) (x – 3y – 5z)(x² + 9y² + 25z² + 3xy – 15yx + 5zx)

⇒ Here the identity used is

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

Comparing this with given condition

(x – 3y – 5z)(x² + 9y² + 25z² + 3xy – 15yx + 5zx)

= x³ + (– 3y)³ + (– 5z)³ – 3(x) (– 3y)(– 5z)

= x³ – 27y³ – 125z³ – 45xyz

(i) 6³ – 9³ + 3³

Here the identity used will be

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

∴ 6³ – 9³ + 3³

⇒ (6 + (– 9) + 3)(6² + (– 9)² + 3² – 6 × – 9 – (– 9)3 – 6 × 3) + 3 × 6 × – 9 × 3

⇒ 0 (36 + 81 + 9 + 54 + 27 – 18) – 486

⇒ – 486

(ii) 16³ – 6³ – 10³

Here the identity used will be

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

∴ 16³ – 6³ – 10³

= (16 – 6 – 10) (16² + (– 6)² + (– 10)² – 16 × – 6 – (– 6)(– 10) – (16)(– 10)) + 3(16) (– 6) (– 10)

= 0 (256 + 36 + 100 + 96 – 60 + 160) + 2880

= 2880

= 2a³ – 3a²b + 2a²c
Highest common factor is a²

Taking a² common in the term – 3a²b + 2a²c

we get,
= 2a³ + a²(– 3b + 2c)
Taking a² common in both the terms
we get,

= a²(2a + (– 3b + 2c))
= a²(2a – 3a + 2c)
2a³ – 3a²b + 2a²c = a²(2a – 3a + 2c)

= 16x + 64x²y

Highest common factor is 16x

Taking x common in the term 16x + 64x²y

we get,

= x(16 + 64xy)

= x(16 + 16(4)xy)

Taking 16 common in the term (16 + 64xy)

we get,

= 16x(1 + 4xy)

16x + 64x²y = 16x(1 + 4xy)

= 10x³ – 25x⁴y
Highest common factor is 5x³

Taking x³common in the term 10x³ – 25x⁴y

we get,

= x³ (10– 25xy)

= x³ (5(2)– 5(2)xy)
Taking 5 common in both the terms

we get,

= 5x³ (2– 5xy)
10x³ – 25x⁴y = 5x³ (2– 5xy)

= xy – xz + ay – az
Taking x common in xy – xz

we get,

= x(y – z) + ay – az

Taking ‘a’ common in ay – az

we get,

= x(y – z) + a(y – z)
Taking (y – z) common in both the terms

we get,
= (y – z)(x + a)
xy – xz + ay – az = (y – z)(x + a)

= p² + pq + pr + qr
Taking p common in p² + pq

we get,

= p(p + q) + pr + qr
Taking r common in pr + qr
we get,
= p(p + q) + r(p + q)

Taking (p + q) common in both the terms

we get,
= (p + q)(p + r)
p² + pq + pr + qr = (p + q)(p + r)

= x² + 2x + 1

Method 1

It is of the form a² + 2ab + b²
Where a = x and b = 1.

Using the identity: (a + b)² = a² + 2ab + b²

We get,

⇒ x² + 2x + 1 = (x + 1)²

Method 2
Here 2x can be elaborated as x + x

we get,

= x² + x + x + 1

Taking x common in x² + x

we get,

= x(x + 1) + x + 1
Taking (x + 1) common in both the terms

we get,
= (x + 1)(x + 1)
= (x + 1)²
x² + 2x + 1 = (x + 1)²

= 9x² – 24xy + 16y²
Given equation can be simplified as,

= (3x)² – 2(3x)(4y) + (4y)²

It is of the form a² – 2ab + b²
Where a = 3x and b = 4y
∴ using the identity (a – b)² = a² – 2ab + b²

We get,
⇒ 9x² – 24xy + 16y² = (3x – 4y)²

9x² – 24xy + 16y² = (3x – 4y)²

= b² – 4
Given equation can be simplified as,

= (b)² – (2)²

It is of the form (p)² – (q)²
where p = b and q = 2
∴ using the identity (p)² – (q)² = (p + q)(p – q)

We get,
⇒ b² – 2² = (b + 2)(b – 2)
b² – 4 = (b + 2)(b – 2)

= 1 – 36x²
Given equation can be simplified as,

= (1)² – (6x)²

It is of the form (p)² – (q)²
Where p = 1 and q = 6x
∴ using the identity: (p)² – (q)² = (p + q)(p – q)

We get,
⇒ (1)² – (6x)² = (1 + 6x)(1 – 6x)

1 – 36x² = (1 + 6x)(1 – 6x)

Method 1
= p² + q² + r² + 2pq + 2qr + 2rp
= (p)² + (q)² + (r)² + 2(p)(q) + 2(q)(r) + 2(r)(p)
It is of the from a² + b² + c² + 2ab + 2bc + 2ca

where a = p, b = q, c = r

Using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

We get,
⇒ p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²

Method 2

= p² + q² + r² + 2pq + 2qr + 2rp
The above equation can be simplified as

= p² + 2pq + q² + r² + 2qr + 2rp
= (p + q)² + r² + 2qr + 2rp (∵ a² + 2ab + b² = (a + b)²)

Taking 2r common in the term 2qr + 2rp

we get,

= (p + q)² + r² + 2r(q + p)
Rearranging the above equation as
= (p + q)² + 2r(q + p) + r²

This is of the from a² + 2ab + b²

Where, a = p + q and b = r
Using the identity: (a + b)² = a² + 2ab + b²

We get,

⇒ p² + q² + r² + 2pq + 2qr + 2rp = ((p + q) + r)²

= (p + q + r)²
p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²

= a² + 4b² + 36 – 4ab + 24b – 12a

Method 1

The above equation can be simplified as :
= (– a)² + (2b)² + (6)² + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
This equation is of the form: p² + q² + r² + 2pq + 2qr + 2rp

where p = – a, q = 2b, r = 6
Using the identity: p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²
We get,
⇒ a² + 4b² + 36 – 4ab + 24b – 12a = (– a + 2b + 6)²

= (– a + 2b + 6)²

= (– 1)²(a – 2b – 6)²

= (a – 2b – 6)²

Method 2

The above equation can be simplified as

= (– a)² + (2b)² + (6)² + 2(– a)(2b) + 2(2b)(6) + 2(– a)(6)
= {(– a)² + 2(– a)(2b) + (2b)²} + (6)² + 2(2b)(6) + 2(– a)(6)
(∵ a² + 2ab + b² = (a + b)²)
(– a + 2b)² + (6)² + 2(2b)(6) + 2(– a)(6)
Taking 2(6) common in term 2(2b)(6) + 2(– a)(6)
= (– a + 2b)² + 2(6)(– a + 2b) + (6)²

This of the form: (p + q)² + 2r(p + q) + r²

Where, p = – a, q = 2b and r =6

Using the identity: (p + q)² = p² + 2pq + q²

We get,

⇒ a² + 4b² + 36 – 4ab + 24b – 12a = (– a + 2b + 6)²

= (– a + 2b + 6)²

= (– 1)²(a – 2b – 6)²

= (a – 2b – 6)²

a² + 4b² + 36 – 4ab + 24b – 12a = (– a + 2b + 6)²

Method 1

The above equation can be simplified as :
= (3x)² + (– y)² + (1)² + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
This equation is of the form: p² + q² + r² + 2pq + 2qr + 2rp

where p =3x, q = – y, r = 1
Using the identity: p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²
We get,
⇒ 9x² + y² + 1 – 6xy + 6x – 2y = (3x – y + 1)²

= (3x – y + 1)²

Method 2

The above equation can be simplified as:

= (3x)² + (– y)² + (1)² + 2(3x)(– y) + 2(3x)(1) + 2(– y)(1)
= {(3x)² + 2(3x)(– y) + (– y)²} + (1)² + 2(3x)(1) + 2(– y)(1)
(∵ p² + 2pq + q² = (p + q)²)
= (3x – y)² + (1)² + 2(3x)(1) + 2(– y)(1)
Taking 2(1) common in term 2(3x)(1) + 2(– y)(1)
= (3x – y)² + (1)² + 2(1)(3x – y)
= (3x – y)² + 2(1)(3x – y) + (1)²

This of the form: (p + q)² + 2r(p + q) + r²

Where, p = 3x, q = – y and r = 1

Using the identity: (a + b)² = a² + 2ab + b²

We get,

⇒ 9x² + y² + 1 – 6xy + 6x – 2y = (3x – y + 1)²
9x² + y² + 1 – 6xy + 6x – 2y = (3x – y + 1)²

Method 1

The above equation can be simplified as :
= (2a)² + (– b)² + (3c)² + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
This equation is of the form: p² + q² + r² + 2pq + 2qr + 2rp

where p =2a, q = – b, r = 3c
Using the identity: p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²
We get,
⇒ 4a² + b² + 9c² – 4ab – 6bc + 12ca = (2a – b + 3c)²

= (2a – b + 3c)²

Method 2

The above equation can be simplified as:

= (2a)² + (– b)² + (3c)² + 2(2a)(– b) + 2(– b)(3c) + 2(3c)(2a)
= {(2a)² + 2(2a)(– b) + (– b)²} + (3c)² + 2(– b)(3c) + 2(3c)(2a)
(∵ p² + 2pq + q² = (p + q)²)

= (2a – b)² + (3c)² + 2(– b)(3c) + 2(3c)(2a)
Taking 2(3c) common in term 2(– b)(3c) + 2(3c)(2a)
= (2a – b)² + (3c)² + 2(3c)(– b + 2a)
= (2a – b)² + 2(3c)(2a – b) + (3c)²

This of the form: (p + q)² + 2r(p + q) + r²

Where, p = 2a, q = – b and r = 3c

Using the identity: (p + q)² = p² + 2pq + q²

We get,

⇒ 4a² + b² + 9c² – 4ab – 6bc + 12ca = (2a – b + 3c)²
4a² + b² + 9c² – 4ab – 6bc + 12ca = (2a – b + 3c)²

Method 1

The above equation can be simplified as :
= (– 5x)² + (2y)² + (3z)² + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
This equation is of the form: p² + q² + r² + 2pq + 2qr + 2rp

where p = – 5x, q = 2y, r = 3z
Using the identity: p² + q² + r² + 2pq + 2qr + 2rp = (p + q + r)²
We get,
⇒ 25x² + 4y² + 9z² – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)²

= (– 5x + 2y + 3z)²

= (– 1)²(5x – 2y – 3z)²

= (5x – 2y – 3z)²

Method 2

The above equation can be simplified as:

= (– 5x)² + (2y)² + (3z)² + 2(– 5x)(2y) + 2(2y)(3z) + 2(3z)(– 5x)
= {(– 5x)² + 2(– 5x)(2y) + (2y)²} + (3z)² + 2(2y)(3z) + 2(3z)(– 5x)
(∵ p² + 2pq + q² = (p + q)²)
= (– 5x + 2y)² + (3z)² + 2(2y)(3z) + 2(3z)(– 5x)
Taking 2(3z) common in term 2(2y)(3z) + 2(3z)(– 5x)
= (– 5x + 2y)² + (3z)² + 2(3z)(2y – 5x)
= (– 5x + 2y)² + 2(3z)(2y – 5x) + (3z)²

This of the form: (p + q)² + 2r(p + q) + r²

Where, p = – 5x, q = 2y and r = 3z

Using the identity: (p + q)² = p² + 2pq + q²

We get,

⇒ 25x² + 4y² + 9z² – 20xy + 12yz – 30zx = (– 5x + 2y + 3z)²

= (– 5x + 2y + 3z)²

= (– 1)²(5x – 2y – 3z)²

= (5x – 2y – 3z)²

25x² + 4y² + 9z² – 20xy + 12yz – 30zx = (5x – 2y – 3z)²

The above equation can be simplified as

= (3x)³ + (4y)³
Both the given terms are perfect cubes, factor using the sum of cubes formula,

(a)³ + (b)³ = (a + b)(a² – ab + b²)

Here a = 3x and b = 4y

⇒ (3x)³ + (4y)³ = (3x + 4y)((3x)² – (3x)(4y) + (4y)²)

= (3x + 4y)(9x² – 12xy + 16y²)
27x³ + 64y³ = (3x + 4y)(9x² – 12xy + 16y²)

The above equation can be simplified as

= (m)³ + (2)³
Both the given terms are perfect cubes, factor using the sum of cubes formula,

(a)³ + (b)³ = (a + b)(a² – ab + b²)
Here a = m and b = 2

⇒ (m)³ + (2)³ = (m + 2)((m)² – (m)(2) + (2)²)

= (m + 2)(m² – 2m + 4)
m³ + 8 = (m + 2)(m² – 2m + 4)

The above equation can be simplified as

= (a)³ + (5)³
Both the given terms are perfect cubes, factor using the sum of cubes formula,

(a)³ + (b)³ = (a + b)(a² – ab + b²)
Here a = a and b = 5

⇒ (a)³ + (5)³ = (a + 5)((a)² – (a)(5) + (5)²)

= (a + 5)(a² – 5a + 25)
a³ + 123 = (a + 5)(a² – 5a + 25)

The above equation can be simplified as

= (2x)³ – (3y)³
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,

(a)³ – (b)³ = (a – b)(a² + ab + b²)
Here a = 2x and b = 3y

⇒ (2x)³ – (3y)³ = (2x – 3y)((2x)² + (2x)(3y) + (3y)²)

= (2x – 3y)(4x² + 6xy + 9y²)
8x³ – 27y³ = (2x – 3y)(4x² + 6xy + 9y²)

The above equation can be simplified as

= (x)³ – (2y)³
Both the given terms are perfect cubes, factor using the subtraction of cubes formula,

(a)³ – (b)³ = (a – b)(a² + ab + b²)
Here a = x and b = 2y

⇒ (x)³ – (2y)³ = (x – 2y)((x)² + (x)(2y) + (2y)²)

= (x – y)(x² + 2xy + 4y²)
x³ – 8y³ = (x – y)(x² + 2xy + 4y²)

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]

x + 3y = 10……(1)

2x + y = 5 …..(2)

From eq(1),

X + 3y = 10

⇒ x = 10 – 3y

Substituting the value of x in eq(2)

2x + y = 5

⇒ 2(10 – 3y) + y = 5

⇒ 20 – 6y + y = 5

⇒ 20 – 5y = 5

⇒ 5y = 20 – 5

⇒ 5y = 15

Substituting value of y in eq(1)

x + 3y = 10

⇒ x + 3×3 = 10

⇒ x + 9 = 10

⇒ x = 10 – 9

⇒ x = 1

Hence,

x = 1, y = 3

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]

2x + y = 1….(1)

3x – 4y = 18….(2)

From eq(1),

2x + y = 1….(1)

⇒ y = 1 – 2x

Substituting the value of y in eq(2)

3x – 4y = 18….(2)

⇒ 3x – 4(1 – 2x) = 18

⇒ 3x – 4 + 8x = 18

⇒ 11x + 4 = 18

⇒ 11x = 18 + 4

⇒ 11x = 22

⟹x =

⇒ x = 2

Substituting the value of x in eq(1),

⇒ 2x + y = 1

⇒ 2×2 + y = 1

⇒ y = 1 – 4

⇒ y = – 3

Hence,

x = 2,y = – 3

[In substitution method, we have to find value of one variable in terms of the other variable from any one of the two given equations . Then, we will substitute this value in the other equation. By doing this, we will get a equation in one variable which can be solved easily]

5x + 3y = 21……(1)

2x – y = 4…….(2)

From eq(2)

2x – y = 4

⟹y = 2x – 4…..(3)

Substituting value of y in eq(1)

5x + 3y = 21

⇒ 5x + 3(2x – 4) = 21

⇒ 5x + 6x – 12 = 21

⇒ 11x = 33

⇒ x =

⇒ x = 3

From eq(3)

y = 2x – 4

⇒ y = 2×3 – 4

⇒ y = 2

Hence,

x = 3, y = 2

In such questions, if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]

Let

= u

= v

Then,

= 9

⇒ u + 2v = 9……(1)

And

= 12

⇒ 2u + v = 12……(2)

From eq(1)

u + 2v = 9

⇒ u = 9 – 2v……(3)

Substituting value of u in eq(2)

2u + v = 12

⇒ 2(9 – 2v) + v = 12

⟹18 – 4v + v = 12

⇒ 18 – 3v = 12

⇒ 3v = 18 – 12

⇒ 3v = 6

⇒ v = 2

⇒ = 2

Substituting value of v in eq(3)

u = 9 – 2v

⇒ u = 9 – 2×2

⇒ u = 5

⇒ = 5

⇒

Hence,

x = ,y =

In such questions , if we start question directly, then it will involve LCM of x and y, and a term of xy, which will be inconvenient for us. So, we will assume 1/x as “u “and 1/y as v”]

+ = 7

⇒ 3u + v = 7……(1)

= 6

⇒ 5u – 4v = 6 ……..(2)

From eq(1)

v = 7 – 3u…..(3)

Substituting the value of v in eq(2)

5u – 4(7 – 3u) = 6

⇒ 5u – 28 + 12u = 6

⇒ 17u = 6 + 28

⇒ 17u = 34

⇒ u = 34/17

⇒ u = 2

⇒ = 2

Substituting the value of u in eq(3)

v = 7 – 3×2

⇒ v = 1

⇒ = 1

⇒ y = 1

Hence, x = , y = 1

Let the two numbers be x and y.

Since, it is given that their sum is 24,

We can write, x + y = 24

For forming second equation, we will use their difference, which is 8.

So, we can write,

x – y = 8

(we can assume any number to be greater. Here, we assumed the greater number is x)

Here, coefficient of y is equal in magnitude and opposite in sign.

So, if we add the above two equations, it will be eliminated.

Adding equation (1)and(2)

We get

2x = 32

⇒ x =

⇒ x = 16

Substituting the value of x in eq(1)

We get

x + y = 24

⇒ 16 + y = 24

⇒ y = 24 – 16

⇒ y = 8

Hence, the two required numbers are 16 and 8.

Let unit digit of the number be x and its ten’s digit be y.

Then the number will be 10×y + x = 10y + x

Sum of its digits is 9.

Therefore, the sum of x and y is 9

x + y = 9……(1)

On reversing the number, ten’s digit will become unit’s digit and unit’s digit will become ten’s digit.

So, after reversing the number,

Its ten’s digit = x and unit’s digit = y

Therefore, the new number formed = 10x + y

New number formed exceeds the twice the original number by 18

⇒ 10x + y = 2(10y + x) + 18

⇒ 10x + y = 20y + 2x + 18

⇒ 8x – 19y = 18…..(2)

From equation(1),

x + y = 9

⇒ x = 9 – y…..(3)

Substituting the value of x in eq(2)

8(9 – y) – 19y = 18

⇒ 72 – 8y – 19y = 18

⇒ – 27y = 18 – 72

⇒ – 27y = – 54

⇒ y = 2

Substituting the value of y in eq(3)

x = 9 – 2

⇒ x = 7

Hence, the original number is 10y + x = 10×2 + 7 = 27

Let Kavi has x apples and Kural has y apples.

According to Kavi’s statement,

We can write

x + 4 = 3(y – 4)

x + 4 = 3y – 12

⇒ x = 3y – 16…..(1)

According to Kural’s statement

y + 26 = 2(x – 26)

y + 26 = 2x – 52…….(2)

Substituting the value of x from eq(1) in eq(2)

y + 26 = 2(3y – 16) – 52

⇒ y + 26 = 6y – 32 – 52

⇒ y + 26 = 6y – 84

⇒ 5y = 110

⇒ y = 22

Substituting value of y in eq(1)

x = 3×22 – 16

⇒ x = 50

Hence, Kavi has 50 apples and Kural has 22 apples.

(i) 2x + 7>15

2x>15 – 7

⇒ 2x>8

⇒ x>

⇒ x>4

(ii) 2x – 4<3

⇒ 2x>3 + 4

⇒ 2x>7

⇒ x>7/2

(iii) 2(x + 7)≤9

⇒ 2x + 14≤9

⇒ 2x≤9 – 14

⇒ 2x≤ – 5

(iv) 3x + 14≥8

⇒ 3x≥8 – 14

⇒ 3x≥ – 6

⇒ x≥

⇒ x≥ – 2

Using identity,

(x + a)(x + b) = x² + (a + b)x + ab

We have,

(x + 2)(x – 1)

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

Using identity,

(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

We have,

(x + 1)(x – 2)(x + 3) = x³ + (1 – 2 + 3)x² + (– 2 – 6 + 3)x – 6

= x³ + 2x² – 5x – 6

(x – y)(x² + xy + y²) is simply the expansion of x³ – y³.

we can also obtain this result by multiplying them

i.e. (x – y)(x² + xy + y²) = x(x² + xy + y²) – y(x² + xy + y²)

= x³ + x²y + xy² – x²y – xy² – y³

= x³ – y³

Suppose (x + p)(x + q) are two factors of x² + 2x – 8.

Then, x² + 2x – 8 = (x + p)(x + q)

= x² + (p + q)x + pq

So, to factorize we have to find p and q, such that pq = – 8 and p + q = 2.

∴ x² + 2x – 8 = (x + 4)(x – 2)

Let the other factor be (x + p)

Then, x² – 6x – 16 = (x + p)(x + 2)

= x² + (p + 2)x + 2p

On comparing the coefficients on both sides, we get,

p + 2 = – 6

⇒ p = – 8

∴ The other factor is (x – 8).

ax² – 5x + c = (2x + 1)(x – 3)

= 2x² – 6x + x – 3

= 2x² – 5x – 3

On comparing the coefficients on both the sides, we get,

a = 2 and c = – 3

x + y = 10

⇒ y = 10 – x substituting this value in x – y = 2

We get,

x – (10 – x) = 2

⇒ 2x – 10 = 2

⇒ 2x = 12

⇒ x = 6

2 – x <5

Adding x on both the sides,

2 <5 + x

⇒ – 3 < x

⇒ x > – 3