Question

if a=x-y, b=y-z, c=z-x then find the value of (a)3+(b)3+(c)3
(a cube+b cube+c cube)​

Answer 1

Answer:

a+b+c = x-y+y-z+z-x = 0

a3 + b3 + C3 - 3abc

= (a+b+c)(a2+b2+c2-ab-bc-ca)

as a+b+c = 0

a3+b3+c3-3abc = 0

so, a3 + b3 + c3 = 3abc

i.e., a3 + b3 + c3

= 3(x-y)(y-z)(z-x)