Question

If a+b+c=0 find the value of (a+b-c)^3+(c+a-b)^3+(b+c-a)^3
Please tell the answers with full explanation ​

Answer 1

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:

ans is 0(zero)

Step-by-step explanation:

Let's simplify step-by-step.

(a+b−c)3+(c+a−b)3+(b+c−a)3

Distribute:

=a3+3a2b+−3a2c+3ab2+−6abc+3ac2+b3+−3b2c+3bc2+−c3+a3+−3a2b+3a2c+3ab2+−6abc+3ac2+−b3+3b2c+−3bc2+c3+−a3+3a2b+3a2c+−3ab2+−6abc+−3ac2+b3+3b2c+3bc2+c3

Combine Like Terms:

=a3+3a2b+−3a2c+3ab2+−6abc+3ac2+b3+−3b2c+3bc2+−c3+a3+−3a2b+3a2c+3ab2+−6abc+3ac2+−b3+3b2c+−3bc2+c3+−a3+3a2b+3a2c+−3ab2+−6abc+−3ac2+b3+3b2c+3bc2+c3

=(a3+a3+−a3)+(3a2b+−3a2b+3a2b)+(−3a2c+3a2c+3a2c)+(3ab2+3ab2+−3ab2)+(−6abc+−6abc+−6abc)+(3ac2+3ac2+−3ac2)+(b3+−b3+b3)+(−3b2c+3b2c+3b2c)+(3bc2+−3bc2+3bc2)+(−c3+c3+c3)

=a3+3a2b+3a2c+3ab2+−18abc+3ac2+b3+3b2c+3bc2+c3

Answer:

=a3+3a2b+3a2c+3ab2−18abc+3ac2+b3+3b2c+3bc2+c3

now you can replace values of a,b and c