# EXPAND THE FOLLOWING USING SUITABLE IDENTIES

To expand the binomial, we will use the identities given below.

(a+b)2 = a2 + 2ab+ b2

(a-b)2 = a2 - 2ab+ b2

a2 - b2 = (a + b) (a - b)

(a+b)3 = a3 + 3a2b + 3ab2 + b3

(a-b)2 = a2 -  + 3a2b + 3ab2 - b3

Expand the following :

Problem 1 :

(3m + 5)2

Solution :

= (3m + 5)2

Using the algebraic identity

(a + b)2 = a2 + 2ab + b2

Here a = 3m and b = 5

(3m + 5)2 = (3m)2 + 2(3m)(5) + 52

= 32m2 + 30m + 25

= 9m2 + 30m + 25

Problem 2 :

(5p - 1)2

Solution :

= (5p - 1)2

Using the algebraic identity

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

Here a = 5p and b = 1

(5p - 1)2 = (5p)2 - 2(5p)(1) + 12

= 52p2 - 10p + 1

= 25p2 - 10p + 1

Problem 3 :

(2n - 1)(2n + 3)

Solution :

= (2n - 1)(2n + 3)

Multiplying the both terms.

= 4n2 + 6n - 2n - 3

= 4n2 + 4n - 3

Problem 4 :

4p2 - 25q2

Solution :

= 4p2 - 25q2

Using the algebraic identity

a2 - b2 = (a + b)(a - b)

4p2 - 25q2 = (2p)2 - (5q)2

Here a = 2p and b = 5q

(2p)2 - (5q)2 = (2p + 5q)(2p - 5q)

Problem 5 :

(3 + m)3

Solution :

= (3 + m)3

Using the algebraic identity

(a + b)3 = a3 + 3a2b + 3ab2 + b3

Here a = 3 and b = m

(3 + m)3 = 33 + 3(3)2(m) + 3(3)(m)2 + m3

= 27 + 27m + 9m2 + m3

= m3 + 9m2 + 27m + 27

Problem 6  :

(2a + 5)3

Solution :

= (2a + 5)3

Using the algebraic identity

(a + b)3 = a3 + 3a2b + 3ab+ b3

Here a = 2a and b = 5

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

= 8a3 + 60a2 + 150a + 125

Problem 7 :

(3p + 4q)3

Solution :

= (3p + 4q)3

Using the algebraic identity

(a + b)3 = a3 + 3a2b + 3b2a + b3

Here a = 3p and b = 4q

(3p + 4q)3 = (3p)3 + 3(3p)2(4q) + 3(3p)(4q)2 + (4q)3

= 27p3 + 108p2q + 144q2p + 64q3

Problem 8 :

(5 - x)3

Solution :

= (5 - x)3

Using the algebraic identity

(a - b)3 = a3 - 3a2b + 3b2a - b3

Here a = 5 and b = x

(5 - x)3 = (5)3 - 3(5)2x + 3(x)2(5) - (x)3

= 125 - 75x + 15x2 - x3

Problem 9 :

(2x - 4y)3

Solution :

= (2x - 4y)3

Using the algebraic identity

(a - b)3 = a3 - 3a2b + 3b2a - b3

Here a = 2x and b = 4y

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

= 8x3 - 48x2y + 96xy2 - 64y3

Problem 10 :

(ab - c)3

Solution :

= (ab - c)3

Using the algebraic identity

(a - b)3 = a3 - 3a2b + 3b2a - b3

Here a = ab and b = c

(ab - c)3 = (ab)3 - 3(ab)2(c) + 3(c)2(ab) - (c)3

= a3b3 - 3a2b2c + 3abc2 - c3

## Evaluate Using Algebraic Identities

Problem 1 :

(52)3

Solution :

= (50 + 2)3

Here a = 50 and b = 2

= 503 + 3(50)2 (2) + 3(50) 22 + 23

= 125000 + 6(2500) + 12(50) + 8

= 125000 + 15000 + 600 + 8

= 140608

Problem 2 :

(104)3

Solution :

= (104)3

= (100 + 4)3

Here a = 100 and b = 4

= (100)3 + 3(100)2(4) + 3(100) 42 + 43

= 1000000 + 120000 + 4800 + 64

= 1124864

Problem 3 :

(48)3

Solution :

= (48)3

= (50 - 2)3

a = 50 and b = 2

= 503 - 3(50)2 (2) + 3(50) 22 - 23

= 125000 - 15000 + 600 - 8

= 110592

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