Question

If a + b + c = 0, then a(b-c)³ + b(c-a)³ + c(a-b)³ =
(A) 1
(B) 0
(C) - 1
(D) none
Please send the full solution of this question.​

Answer 1

Answer:

b). 0

Step-by-step explanation:

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=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

3xyz=3∗(−2c)∗(−2b)∗(−2a)=−24abc

Hope this helped!

Answer 2

Answer:

0

Step-by-step explanation:

If a + b + c = 0, then a(b-c)³ + b(c-a)³ + c(a-b)³ = 0