Question

x+y+z=0 show that x^3+y^3+z^3=3xyz​

Answer 1

Answer:

Proved

Step-by-step explanation:

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Answer 2

$\LARGE{\bf{\underline{\underline{\black{GIVEN:-}}}}}$

• $\sf x+y+z=0$

$\LARGE{\bf{\underline{\underline{\black{TO:FIND:-}}}}}$

• $\sf x^3+y^3+y^3=3xyz$

$\LARGE{\bf{\underline{\underline{\black{SOLUTION:-}}}}}$

Given that:

$\sf x+y+z=0$

Transpose z to the RHS.

$\sf x+y =-z$______(1)

Cube on both sides.

$\sf \to (x+y)^3=z^3$

Expand the LHS.

$\to \sf x^3+y^3+3xy(x+y)=-z^3$___________(2)

Substitute (1) in (2).

$\sf \to x^3+y^3+3xy(-z)=-z^3$

$\sf \to x^3+y^3-3xyz=-z^3$

Transpose -3yz to RHS and -z³ to LHS.

$\boxed{\sf{\blue{x^3+y^3+z^3=3xyz}}}$

Thus proved.