Complex Numbers 5 Question 19
20.
If $x=a+b, y=a \alpha+b \beta, z=a \beta+b \alpha$, where $\alpha, \beta$ are complex cube roots of unity, then show that $x y z=a^{3}+b^{3}$.
$(1979,3 M)$
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Solution:
- Since, $\alpha, \beta$ are the complex cube roots of unity.
$\therefore$ We take $\alpha=\omega$ and $\beta=\omega^{2}$.
Now, $\quad x y z=(a+b)(a \alpha+b \beta)(a \beta+b \alpha)$
$=(a+b)\left[a^{2} \alpha \beta+a b\left(\alpha^{2}+\beta^{2}\right)+b^{2} \alpha \beta\right]$
$=(a+b)\left[a^{2}\left(\omega \cdot \omega^{2}\right)+a b\left(\omega^{2}+\omega^{4}\right)+b^{2}\left(\omega \cdot \omega^{2}\right)\right]$
$=(a+b)\left(a^{2}-a b+b^{2}\right) \quad\left[\because 1+\omega+\omega^{2}=0\right.$ and $\left.\omega^{3}=1\right]$
$=a^{3}+b^{3}$
