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

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Solution

## 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,$\begin{array}{rcl}xyz& =& \left(a+b\right)\left(a\alpha +b\beta \right)\left(a\beta +b\alpha \right)\\ & =& \left(a+b\right)\left({a}^{2}\alpha \beta +ab{\alpha }^{2}+ab{\beta }^{2}+{b}^{2}\alpha \beta \right)\\ & =& \left(a+b\right)\left({a}^{2}\alpha \beta +ab\left({\alpha }^{2}+{\beta }^{2}\right)+{b}^{2}\alpha \beta \right)\end{array}$Now substitute values of $\alpha ,\beta$So,$\begin{array}{rcl}xyz& =& \left(a+b\right)\left({a}^{2}\left(\omega {\omega }^{2}\right)+ab\left({\omega }^{2}+{\omega }^{4}\right)+{b}^{2}\left(\omega {\omega }^{2}\right)\right)\\ & =& \left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)\mathbf{\left[}\mathbf{\because }\mathbf{}{\mathbf{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{\right]}\\ & =& {a}^{3}+{b}^{3}\end{array}$Hence, the correct answer is ${a}^{3}+{b}^{3}$.

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