Factorize:
a3(b – c)3 + b3(c – a)3 + c3(a – b)3
We have,
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
We know that,
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)]
Also,
If, (x + y + z) = 0
Then (x3 + y3 + z3) = 3xyz
= [a (b – c)]3 + [b (c – a)]3 + [c (a – b)]3
Since,
a (b – c) + b (c – a) + c (a – b) = ab – ac + bc – ba + ca – bc = 0
So,
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
= 3a (b – c) b (c – a) c (a – b)
= 3abc (a – b) (b – c) (c – a)