Question

8. Factorise each of the following
(1) 8a^3+ b^3 +12a+b + 6ab^2
(ii) 8a^3 - b^3 - 12a+b + 6ab^2
(iii) 1 - 64a^3 - 12a + 48a^2
(iv) 8p^3-12/5p^2+6/25p-1/125​

Answer 1

Answer:

I).8a^3+b^3+12a+b +6ab^2

Step-by-step explanation:

=(2a)^3+(b)^3+12a^2b+6ab^2

=(2a)^3+(b)^3+6ab (2a+b)

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

using of(x+y)^3=x^3+y^3+3xy(x+y)

putting,x=2a, y=b

=(2a+b)^3

=(2a+b) (2a+b) (2a+b)

Answer:

ii).8a^3-b^3-12a+b+6ab^2

Step-by-step explanation:

=(2a-b)(4a²+2ab+b²)-6ab(2a-b)

=(2a-b)(4a²+2ab+b²-6ab)

=(2a-b)(4a²-4ab+b²)

=(2a-b){(2a)²-2×2a×b+(b)²}

=(2a-b)(2a-b)²

=(2a-b)(2a-b)(2a-b)

Answer:

iii).1−64a^3−12a+48a^2

=(1)^3−(4a) ^3−3(1)^2(4a)+3(1)(4a)^2

Suitable identities is x^3-y^3-3x^2y+3xy^2=(x-y)^3

•°•(1)^3-(4a)^3-3(1)^2(4a)+3(1) (4a)^2

=(1-4a)^3

Answer 2

Step-by-step explanation:

thanks for free points

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