Q18 of 34 Page 252

Factorize: (p-q)3 + (q-r)3 + (r-p)3

We know that –

a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca).


here if a + b + c = 0


a3 + b3 + c3 = 3abc.


So, (p-q)3 + (q-r)3 + (r-p)3 = 3(p-q)(q-r)(r-p) {since (p-q) + (q-r) + (r-p) = 0}


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