Question

Please solve this whole sum​

Answer 1

Answer:

Solutions:

(i) 8a3+b3+12a2b+6ab2

Solution:

The expression, 8a3+b3+12a2b+6ab2 can be 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

Solution:

The expression, 8a3–b3−12a2b+6ab2 can be 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

Solution:

The expression, 27 – 125a3 – 135a + 225a2 can be 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

Solution:

The expression, 64a3–27b3–144a2b+108ab2 can be 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) 27p3 – 1/216 − 9/2p2+ 1/4p

Solution:

The expression, 27p3 –1/216 − 9/2p2+ 1/4p can be written as (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

27p3 – 1/216 − 9/2p2+ 1/4p = (3p)3–(1/6)3–3(3p)2(1/6)+3(3p)(1/6)2

= (3p–(1/6))3

= (3p–(1/6))(3p–(1/6))(3p–(1/6))

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Answer 2

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