Question

If $\alpha$ & $\beta$ are complex cube rout of unity $x=a+b$ ,$y=\alpha a+\beta b$ ,$z=a\beta +b\alpha$ then show $xyz=a^3+b^3$

Answer 1

Consider the problem

Let,

We denote complex cube root of unity as $r$ and $r^2$

where

$1+r+r^2=0$

Also,

$r^3=1$

Now,

$\alpha =r$ and $\beta =r^2$

$x=a+b$

$\begin{array}{c}y=a\alpha +b\beta =ar+b{r}^2\\ \\ z=a\beta +b\alpha =a{r}^2+br\\ \\ xyz=\left(a+b\right)\left(ar+b{r}^2\right)\left(a{r}^2+br\right)\\ \\ =\left(a+b\right)\left(a^2{r}^3+ab{r}^2+ab{r}^4+b^2{r}^3\right)\\ \\ =\left(a+b\right)\left(a^2{r}^3+ab{r}^2+ab{r}^3.r+b^2{r}^3\right)\end{array}$

As $r^3=1$

Then,

$\begin{array}{c}xyz=\left(a+b\right)\left(a^2+ab{r}^2+abr+b^2\right)\\ \\ =\left(a+b\right)\left(a^2+ab\left(r^2+r\right)+b^2\right)\end{array}$

As $1+r+r^2=0$

$r+r^2=-1$

Then,

$\begin{array}{c}=\left(a+b\right)\left(a^2-ab+b^2\right)\\ \\ xyz=a^3-b^3\end{array}$

Hence, proved.