Question

If x+y+z=0 then prove that x^3+y^3+z^3=3xyz

Answer 1

Answer:

x^3+y^3+z^3-3xyz=(X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)

NOW PUT

X+Y+Z=0

RHS BECOMES 0

WE GET

x^3+y^3+z^3=3xyz

Step-by-step explanation:

Answer 2

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

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