Question

please answer fast fast​

Answer 1

Answer:

Here is your answer. Hope it helps you. Mark as Brainliest please

Answer 2

Given that,

$\tt{x + y + z = 0}$

We need to find :

$\tt{The\ value\ of\ x\³ + y\³ + z\³.}$

Solution :-

$\texttt{We know the identity;}$

$\bold{\displaystyle{x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] }}$

$\texttt{We are given that x + y + z = 0, so, if we substitute} \\\texttt{this value in the RHS, we get :}$

$\bold{\displaystyle{x^3+y^3+z^3-3xyz=\frac{1}{2}*(0)*[(x-y)^2+(y-z)^2+(z-x)^2] }}$

$\texttt{We know that, anything multiplied with 0 is 0.}$

$\bold{\displaystyle{x^3+y^3+z^3-3xyz=0}}\\\bold{\displaystyle{\implies x^3+y^3+z^3=3xyz }}$

$\underline{\texttt{So, the answer is }\boxed{\sf{(a)\ 3xyz}}.}$