Question

Prove that:
x^3 + y^3 + ^3 = 3xyz,
if (x + y + z) = 0.

Answer 1

Step-by-step explanation:

Given x+y+z=0

⟹x+y=−z

Cubing on both sides

(x+y)

3

=(−z)

3

⟹x

3

+y

3

+3x

2

y+3xy

2

=−z

3

⟹x

3

+y

3

+3xy(x+y)=−z

3

⟹x

3

+y

3

+3xy(−z)=−z

3

⟹x

3

+y

3

−3xyz=−z

3

⟹x

3

+y

3

+z

3

=3xyz

Answer 2

$x + y + z = 0 = > x + y = - z \\ = > (x + y)^{3} = ( - z )^{3} \\ = > x^{3} + y ^{3} + 3xy \: (x + y) = ( - z) = - {z} \\ = > x^{3} + y ^{3} + 3xy ( - z ) = - {z}^{3} \\ = > x^{3} + y ^{3} - 3xyz = 3xyz$

$Hence,(x + y + z) = 0 = > ( {x}^{3} {y}^{3} {z}^{ 3} ) = 3xyz$