If 
find the value of 
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
Given,
x + y + 4 = 0
We have,
(x3 + y3 – 12xy + 64)
= (x)3 + (y)3 + (4)3 – 3 (x) (y) (4) = 0
18