Question

factorise : [a+b] whole cube minus 8

Answer 1

I think your question is (a+b)^3-8
hence (a+b)^3-(2) ^3
according to formula a^3-b^3=(a-b)(a^2+b^2+ab)
so, (a+b-2){(a+b)^2+4+2 (a+b)}

Answer 2

Answer:

(a+b)³-8

(a+b)³-8= (a+b-2)(a²+b²+2ab+2a+2b+4)

Step-by-step explanation:

Given (a+b)³-8

= (a+b)³-2³

/* By algebraic identity:

$\boxed {x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2}}$*/

= [(a+b)-2][(a+b)²+(a+b)2+2²]

=(a+b-2)(a²+b²+2ab+2a+2b+4)

Therefore,

(a+b)³-8

= (a+b-2)(a²+b²+2ab+2a+2b+4)

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