Question -
Answer -
(i) 8a3+b3+12a2b+6ab2
The expression, 8a3+b3+12a2b+6ab2 canbe written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a3+b3+12a2b+6ab2 =(2a)3+b3+3(2a)2b+3(2a)(b)2
= (2a+b)3
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)3 = x3+y3+3xy(x+y)is used.
(ii) 8a3–b3–12a2b+6ab2
The expression, 8a3–b3−12a2b+6ab2 canbe written as (2a)3–b3–3(2a)2b+3(2a)(b)2
8a3–b3−12a2b+6ab2 =(2a)3–b3–3(2a)2b+3(2a)(b)2
= (2a–b)3
= (2a–b)(2a–b)(2a–b)
Here, the identity,(x–y)3 = x3–y3–3xy(x–y)is used.
(iii) 27–125a3–135a+225a2
The expression, 27–125a3–135a +225a2 canbe written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2
27–125a3–135a+225a2 =
33–(5a)3–3(3)2(5a)+3(3)(5a)2
= (3–5a)3
= (3–5a)(3–5a)(3–5a)
Here, the identity, (x–y)3 = x3–y3-3xy(x–y)is used.
(iv) 64a3–27b3–144a2b+108ab2
The expression, 64a3–27b3–144a2b+108ab2canbe written as (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2
64a3–27b3–144a2b+108ab2=
(4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2
=(4a–3b)3
=(4a–3b)(4a–3b)(4a–3b)
Here, the identity, (x – y)3 = x3 –y3 – 3xy(x – y) is used.
(v) 7p3–(1/216)−(9/2) p2+(1/4)p
The expression, 27p3–(1/216)−(9/2) p2+(1/4)p
can be written as (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2
27p3–(1/216)−(9/2) p2+(1/4)p =
(3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2
= (3p–16)3
= (3p–16)(3p–16)(3p–16)