Question

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Answer 1

(a+b)³-a-b=(a+b)(a+b+1)(a+b-1)

Factorisation of (a+b)³-a-b

= (a+b)(a+b)²-1(a+b)

Take (a+b) common, we get

= (a+b)[(a+b)²-1]

= (a+b)[(a+b)²-1²]

= (a+b)[(a+b)+1][(a+b)-1]

/* By algebraic identity */

= (a+b)(a+b+1)(a+b-1)

Therefore,

(a+b)³-a-b=(a+b)(a+b+1)(a+b-1)

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