Question

The value of [(x-y)^3 + (y-z)^3 + (z-x)^3]/[(9(x-y)(y-z)(z-x)] is equal to: a.) 1 b.) 0 c.) 1/3 d.) 1/9

Answer 1

let x-y =a
y-z =b
z-x = c
now
a+b+c =0
hence you can proceed

Answer 2

do you know the identity if a+b+c =0

a^3 + b^3 + c^3 = 3abc

Answer 3

I was only familiar with square of binomials, stumbled upon that identity for the first time...

Answer 4

if
x-y =a
y-z =b
z-x = c then we have a+b+c=0
now the given question translates to
(a^3 + b^3 + c^3 ) / 9abc = 3abc/9abc =1/3

Answer 5

That helped, I just solved the problem... Thanks a lot! I have more follow up questions..

Answer 6

need not mention..