Question

please answer fast fast and step by step​

Answer 1

Given :-

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

To find :-

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

Solution :-

$\texttt{We know the identity ;}$

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

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

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

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

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

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

$\texttt{So, the answer is:}$

$\boxed{\boxed{\boxed{\boxed{\boxed{\sf{(a)\ 3xyz}}}}}}$

Answer 2

Answer:

answer of your following question is 3 x y z if you bar related my answer to give explanation