Find the product:
(x + y – z)(x2 + y2 + z2 – xy + yz + zx)
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
Using this, we get
= [x + y + (-z)] [(x)2 + (y)2 + (-z)2 – (x) (y) – (y) (-z) – (-z) (x)]
= x3 + y3 – z3 + 3xyz