Question

If $x=a+b,y=a\alpha +b\beta \mathrm{and}\mathrm{z}=a\beta +\mathrm{b\alpha }$, where $\alpha ,\beta$ not equal $1$are cube roots of unity, then $xyz$ equals

Answer 1

Find the value of $xyz$:

Given,

$x=a+b,y=a\alpha +b\beta \mathrm{and}\mathrm{z}=a\beta +\mathrm{b\alpha }$, and $\alpha ,\beta$ are cube roots of unity.

Then,

$\alpha =\omega ,\beta ={\omega }^2$

Now,

$\begin{array}{rcl}xyz& =& (a+b)(a\alpha +b\beta )(a\beta +b\alpha )\\ & =& (a+b)\left(a^2\alpha \beta +ab{\alpha }^2+ab{\beta }^2+b^2\alpha \beta \right)\\ & =& (a+b)\left(a^2\alpha \beta +ab({\alpha }^2+{\beta }^2)+b^2\alpha \beta \right)\end{array}$

Now substitute values of $\alpha ,\beta$

So,

$\begin{array}{rcl}xyz& =& (a+b)\left(a^2\right(\omega {\omega }^2)+ab({\omega }^2+{\omega }^4)+b^2(\omega {\omega }^2\left)\right)\\ & =& (a+b)(a^2-ab+b^2)\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& a^3+b^3\end{array}$

Hence, the correct answer is $a^3+b^3$.