Question

Factorise a³(b-c)³+b³(c-a)³+c³(a-b)³

Answer 1

A³(b - c)³ + b³(c - a)³ + c³(a - b)³

= {a(b - c)}³ + {b( c - a)}³ + {c(a - b)}³

we see that,
a(b - c) + b(c - a) + c ( a - b) = 0
so, we can use ,
x³ + y³ + z³ = 3xyz when, (x + y + z) = 0

hence,
{a(b - c)}³ + {b(c - a)}³ + {c(a - b)}³ = 3abc(b - c)(c -a )(a - b)

hope this helps you out. ....