Question

If $x + y + z = 0$, show that $x^3 + y^3 + z^3 =3xyz$ -

[Explanation MANDATORY]

Answer 1

Answer:

THAT'S TRUE.

Step-by-step explanation:

CONSIDERING X, Y AND Z =0;

0³+0³+0³=3(0)(0)(0) = 0

HOPEFULLY IT HELPS! PLEASE MARK AS BRAINLIEST!

Answer 2

Answer:

$\huge \blue \star \underbrace \pink{answer}$

Given

$x + y + z = 0$

To prove,

$x {}^{3} + y {}^{3} + z {}^{3} = 3xyz$

using the formula,

$x { }^{3} + y {}^{3} + z {}^{3} - 3xyz = (x + y + z) (x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - xz)$

substitute the value for x+y+z as given in the question,

$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = 0(x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - xz)$

$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = 0$

now, bringing (-3xyz) to the Right Hand Side, we get,

$x {}^{3} + y {}^{3} + z {}^{3} = 3xyz$

Step-by-step explanation:

⚡️☺️✔️✔️

Hope it helps u

$\huge \bf{ \blue{♡♡♡}}$