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 = 2y + 6
Or, x – 2y – 6 = 0
We have,
(x3 – 8y3 – 36xy – 216)
= (x3 – 8y3 – 216 – 36xy)
= (x)3 + (-2y)3 + (-6)3 – 3 (x) (-2y) (-6)
= (x – 2y – 6) [(x)2 + (-2y)2 + (-6)2 – (x) (-2y) – (-2y) (-6) – (-6) (x)]
= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 8x)
= 0 (x2 + 4y2 + 36 + 2xy – 12y + 6x)
= 0
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