Question

if a + b + c equal to zero then find the value of (a+b+c)^3 + (c+a-b)^3 + (b+c-a)^3

Answer 1

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)

If x + y + z = 0,

x3+y3+z3−3xyz=0=>x3+y3+z3=3xyzx3+y3+z3−3xyz=0=>x3+y3+z3=3xyz

Assume:

x = a + b - c = -2c

y = a + c - b = -2b

z = b + c - a = -2a

(Since a+b+c=0, a+b=−c, b+c=−a, a+c=−b)

We see x + y + z = a + b + c =0.

(a+b−c)3+(c+a−b)3+(b+c−a)3=x3+y3+z3=3xyz(a+b−c)3+(c+a−b)3+(b+c−a)3=x3+y3+z3=3xyz

3xyz=3∗(−2c)∗(−2b)∗(−2a)=−24abc3xyz=3∗(−2c)∗(−2b)∗(−2a)=−24abc

Hope this helped!

Answer 2

it might be 0

as if it is multiplied by 3 the answer remains 0