Q24 of 34 Page 252

Factorize: a3 – b3 + 1 + 3ab.

a3 – b3 + 1 + 3ab

= a3 + (–b)3 + 13 – 3{1 × a × (–b)}


= {a + (–b) + 1} {a2 + (–b)2 + 12– a(–b) – (–b)1 – 1a}


using identity {a3 + b3 + c3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)} + (a-b + 1)( a2 + b2 + 1 + ab + b – a)


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