Question

If $\omega$ is an imaginary cube root of unity, then the value of the expression
$1(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+....+(n-1)(n-\omega )(n-{\omega }^2)$ is

Answer 1

$=\sum (n-1)(n-\omega )(n-{\omega }^2)$ ......$\left(1\right)$

We know $a^3-b^3=(a-b)(a-b\omega )(a+{\omega }^2)$ ......$\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$ we get

$b=-1$

$\therefore \sum (n-1)(n-1\times \omega )(n-1\times {\omega }^2)$

$=\sum n^3-1^3$

$=\sum n^3-\sum 1$

$={\left(\frac{n(n+1)}{2}\right)}^2-n$

$=\frac{n^2(n^2+2n+1)}{4}-n$

$=\frac{n(n^3+2n^2+n-4)}{4}$ on simplification