Question 15 Marks
If $x=a+b, y=\alpha a+\beta b$ and $z=a \beta+b \alpha$, where $\alpha$ and $\beta$ are complex cube roots of unity, show that $x y z=a^3+b^3$.
Answer
View full question & answer→$x=a+b, y=\alpha a+\beta b, z=a \beta+b \alpha$
$\alpha$ and $\beta$ are the complex cube roots of unity.
$ \therefore \alpha=\frac{-1+i \sqrt{3}}{2} \text { and } \beta=\frac{-1-i \sqrt{3}}{2}$
$\therefore \quad \alpha \beta=\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right)$
$=\frac{(-1)^2-(\mathrm{i} \sqrt{3})^2}{4}$
$=\frac{1-(-1)(3)}{4}$
$=\frac{1+3}{4}$
$\ldots\left[\because \mathrm{i}^2=-1\right]$
$\therefore \quad \alpha \beta=1$
Also, $\alpha+\beta=\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2}$
$=\frac{-1+\mathrm{i} \sqrt{3}-1-\mathrm{i} \sqrt{3}}{2}=\frac{-2}{2}$
$ \therefore \quad \alpha+ \beta=-1$
$\therefore \quad x y z =(a+b)(\alpha a+\beta b)(a \beta+b \alpha)$
$ =(a+b)\left(\alpha \beta a^2+\alpha^2 a b+\beta^2 a b+\alpha \beta b^2\right)$
$ =(a+b)\left[1 .\left(a^2\right)+\left(\alpha^2+\beta^2\right) a b+1 .\left(b^2\right)\right]$
$ =(a+b)\left\{a^2+\left[(\alpha+\beta)^2-2 \alpha \beta\right] a b+b^2\right\}$
$ =(a+b)\left\{a^2+\left[(-1)^2-2(1)\right] a b+b^2\right\}$
$ =(a+b)\left[a^2+(1-2) a b+b^2\right]$
$ =(a+b)\left(a^2-a b+b^2\right)=a^3+b^3$
$\alpha$ and $\beta$ are the complex cube roots of unity.
$ \therefore \alpha=\frac{-1+i \sqrt{3}}{2} \text { and } \beta=\frac{-1-i \sqrt{3}}{2}$
$\therefore \quad \alpha \beta=\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}\right)\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}\right)$
$=\frac{(-1)^2-(\mathrm{i} \sqrt{3})^2}{4}$
$=\frac{1-(-1)(3)}{4}$
$=\frac{1+3}{4}$
$\ldots\left[\because \mathrm{i}^2=-1\right]$
$\therefore \quad \alpha \beta=1$
Also, $\alpha+\beta=\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2}$
$=\frac{-1+\mathrm{i} \sqrt{3}-1-\mathrm{i} \sqrt{3}}{2}=\frac{-2}{2}$
$ \therefore \quad \alpha+ \beta=-1$
$\therefore \quad x y z =(a+b)(\alpha a+\beta b)(a \beta+b \alpha)$
$ =(a+b)\left(\alpha \beta a^2+\alpha^2 a b+\beta^2 a b+\alpha \beta b^2\right)$
$ =(a+b)\left[1 .\left(a^2\right)+\left(\alpha^2+\beta^2\right) a b+1 .\left(b^2\right)\right]$
$ =(a+b)\left\{a^2+\left[(\alpha+\beta)^2-2 \alpha \beta\right] a b+b^2\right\}$
$ =(a+b)\left\{a^2+\left[(-1)^2-2(1)\right] a b+b^2\right\}$
$ =(a+b)\left[a^2+(1-2) a b+b^2\right]$
$ =(a+b)\left(a^2-a b+b^2\right)=a^3+b^3$