Question

If a+b+c=0 then what will value of
(a+b-c)(b+c-a)(c+a-b) will be?

Answer 1

Here is your answer.
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)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!
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