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Simplify: 

i. (2a + b)– (2a – b)3

ii. (3r – 2k)+ (3r + 2k)3

iii. (4a – 3)3 – (4a + 3)3

iv. (5x – 7y)3 + (5x + 7y)3

1 Answer

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Best answer

i. (2a + b)– (2a – b)3

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

= a+ 3a2b + 3ab2 + b3, (a – b)3 = a– 3a2b + 3ab2 – b3

= (8a3 + 12a2b + 6ab2 + b3) – (8a3 – 12a2b + 6ab2 – b3

= 8a3 + 12a2b + 6ab2 + b3 – 8a+ 12a2b – 6ab2 + b

= 8a3 – 8a3 + 12a2b + 12a2b + 6ab2 – 6ab2 + b3 + b3 

= 24a2b + 2b3

ii. (3r – 2k)3 + (3r + 2k)3 

= [(3r)3 – 3(3r)2(2k) + 3(3r)(2k)2 – (2k)3] + [(3r)3 + 3(3r)2(2k) + 3(3r)(2k)2 + (2k)3] … [(a – b)3 = a3 – 3a2b + 3ab2 – b3, (a + b)3= a+ 3a2b + 3ab2 + b3

= (27r3 – 54r2k + 36rk2 – 8k3) + (27r+ 54r2k + 36rk2 + 8k3

= 27r3 – 54r3k + 36rk2 – 8k3 + 27r3 + 54r2k + 36rk2 + 8k3

= 27r3 + 27r3 – 54r2k + 54r2k + 36rk2 + 36rk– 8k3 + 8k3 

= 54r3 + 72rk2

iii. (4a – 3)3 – (4a + 3)3 

= [(4a)3 – 3(4a)2 (3) + 3(4a)(3)2 – (3)3] – [(4a)3 + 3(4a)2(3) + 3(4a)(3)2 + (3)3] … [(a – b)3 = a3 – 3a2b + 3ab2 – b3, (a + b)3 = a3 + 3a2b + 3ab2 + b3

= (64a3 – 144a2 + 108a – 27) – (64a3 + 144a2 + 108a + 27) 

= 64a3 – 144a+ 108a – 27 – 64a3 -144a2 – 108a – 27 

= 64a3 – 64a3 – 144a2 – 144a2 + 108a – 108a – 27 – 27 

= -288a2 – 54

v. (5x – 7y)3 + (5x + 7y)3 

= [(5x)3 – 3(5x)2(7y) + 3(5x)(7y)2 – (7y)3] + [(5x)3 + 3(5x)2 (7y) + 3(5x) (7y)2 +(7y)3] … [(a – b)3 = a– 3a2b + 3ab2 – b3, (a + b)3 = a+ 3a2b + 3ab+ b3

= (125x– 525x2y + 735xy– 343y3) + (125x3 + 525x2y + 735xy2 + 343y3

= 125x3 – 525x2y + 735xy2 – 343y3 + 125x3 + 525x2y + 735xy2 + 343y3 

= 125x3 + 125x3 – 525x2y + 525x2y + 735xy2 + 735xy– 343y3 + 343y3 

= 250x3 + 1470xy2

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