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# $\alpha ,\beta$ are complex cube roots of unity and $x=a+b$, $y=a\alpha +b\beta$, $z=a\beta +b\alpha$ then ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$=A. $0$B. $3ab$C. $3({{a}^{3}}-{{b}^{3}})$D. $3{{(a-b)}^{3}}$

Last updated date: 15th Jul 2024
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Answer
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Hint: Assume $\alpha =\omega ,\beta ={{\omega }^{2}}$. Substitute the values in $x.y$and $z$. Then add $x+y+z$ and simplify. Then use the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$. Then substitute the value of $x+y+z$ in the formula. You will get the answer. Try it.

Complete step-by-step answer:
$x=a+b$
$y=a\omega +b{{\omega }^{2}}$
$z=a{{\omega }^{2}}+b\omega$

So $x+y+z=a+b+a\omega +b{{\omega }^{2}}+a{{\omega }^{2}}+b\omega$
$x+y+z=a(1+\omega +{{\omega }^{2}})+b(1+\omega +{{\omega }^{2}})$

We know $1+\omega +{{\omega }^{2}}=0$.
$x+y+z=0$

We know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.
So here$x+y+z=0$.

So ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(0)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$
${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$
So we know the value of$x,y$ and $z$.

So substituting the values we get,
${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)(a\omega +b{{\omega }^{2}})(a{{\omega }^{2}}+b\omega )$
Simplifying we get,
\begin{align} & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3(a+b)({{a}^{2}}{{\omega }^{3}}+ab{{\omega }^{3}}+ab{{\omega }^{4}}+{{b}^{2}}{{\omega }^{3}}) \\ & {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}}) \\ \end{align}
${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})$

So we get the value of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3({{a}^{3}}+{{b}^{3}})$.

Note: Read the question carefully. Also, while simplifying, don't make any mistake. Take utmost care of the sign. Do not jumble while simplifying. Solve it step by step. You must know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.