Let us factorise the following algebraic expressions:
p3 (q – r)3 + q3 (r – p)3 + r3 (p – q)3
p3 (q – r) 3 + q3 (r – p) 3 + r3 (p – q) 3
⇒ (pq – pr) 3 + (qr – pq) 3 + (pr – qr) 3
Let us consider pq – pr = a, qr – pq = b, pr – qr = c
a + b + c = pq – pr + qr – pq + pr – qr = 0
Since a + b + c = 0, hence
a3 + b3 + c3 = 3abc
⇒ p3 (q – r) 3 + q3 (r – p) 3 + r3 (p – q) 3 = 3pqr(q - r) (r – p) (p – q)
3pqr (q - r) (r – p) (p – q)
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