Question

Let $x=\alpha +\beta ,y=\alpha \omega +\beta {\omega }^2,z=\alpha {\omega }^2+\beta \omega ,\omega$ is an imaginary cube root of unity. Product of $xyz$ is.

• A. ${\alpha }^2+{\beta }^2$
• B. ${\alpha }^2-{\beta }^2$
• C. ${\alpha }^3+{\beta }^3$
• D. ${\alpha }^3-{\beta }^3$

Answer 1

Answer: (C)

Given, $x=\alpha +\beta ,y=\alpha \omega +\beta {\omega }^2$ and $z=\alpha {\omega }^2+\beta \omega$
Also, ${\omega }^3=1$
Now, $xyz=(\alpha +\beta )\left(\alpha \omega +\beta {\omega }^2\right)\left(\alpha {\omega }^2+\beta \omega \right)$
$xyz=\left[{\alpha }^2\omega +\alpha \beta {\omega }^2+\alpha \beta \omega +{\beta }^2{\omega }^2\right]\left[\alpha {\omega }^2+\beta \omega \right]$
$xyz={\alpha }^3{\omega }^3+{\alpha }^2\beta {\omega }^2+{\alpha }^2\beta {\omega }^4+\alpha {\beta }^2{\omega }^3+{\alpha }^2\beta {\omega }^3+\alpha {\beta }^2{\omega }^2+\alpha {\beta }^2{\omega }^4+{\beta }^3{\omega }^3$
$xyz={\alpha }^3+{\alpha }^2\beta {\omega }^2+{\alpha }^2\beta \omega +\alpha {\beta }^2+{\alpha }^2\beta +\alpha {\beta }^2{\omega }^2+\alpha {\beta }^2\omega +{\beta }^3$
$\left[\because {\omega }^4=\omega \right]$
$xyz={\alpha }^3+{\beta }^3+{\alpha }^2\beta \left(1+\omega +{\omega }^2\right)+\alpha {\beta }^2\left(1+\omega +{\omega }^2\right)$
$xyz={\alpha }^3+{\beta }^3+0+0$
$\left[\because 1+\omega +{\omega }^2=0\right]$
$xyz={\alpha }^3+{\beta }^3$