Factorize:
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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)