If w is a complex cube root of unity then show that (2-w)(2-w^2)(2-w^10)(2-w^11)=49?

Aug 7, 2018

Kindly refer to the Explanation.

Explanation:

Since, $w \in \mathbb{C} , w = \sqrt[3]{1} , \text{ we have, } {w}^{3} - 1 = 0$.

$\therefore \left(w - 1\right) \left({w}^{2} + w + 1\right) = 0 , \mathmr{and} , {w}^{2} + w = - 1. \ldots . . \left(\star\right)$.

$\text{The Expression} = \left\{\left(2 - w\right) \left(2 - {w}^{2}\right)\right\} \left\{\left(2 - {w}^{10}\right) \left(2 - {w}^{11}\right)\right\}$,

$= \left\{4 - 2 \left(w + {w}^{2}\right) + {w}^{3}\right\} \left\{4 - 2 \left({w}^{10} + {w}^{11}\right) + {w}^{21}\right\}$,

$= \left\{4 - 2 \left(- 1\right) + 1\right\} \left\{4 - 2 {w}^{9} \left(w + {w}^{2}\right) + {1}^{7}\right\}$,

$= \left(7\right) \left\{4 - 2 \left({1}^{3}\right) \left(- 1\right) + 1\right\}$,

$= \left(7\right) \left(7\right)$,

$= 49$, as desred!

Aug 8, 2018

$\omega$ being imaginary cube root of unity,we know
${\omega}^{3} = 1 \mathmr{and} 1 + \omega + {\omega}^{2} = 0$

$\text{Now the given expression}$

$= \left(2 - w\right) \left(2 - {w}^{2}\right) \left(2 - {w}^{10}\right) \left(2 - {w}^{11}\right)$

$= \left(2 - w\right) \left(2 - {w}^{2}\right) \left(2 - {\left({w}^{3}\right)}^{3} \cdot w\right) \left(2 - {\left({w}^{3}\right)}^{3} \cdot {w}^{2}\right)$

$= \left(2 - w\right) \left(2 - {w}^{2}\right) \left(2 - w\right) \left(2 - {w}^{2}\right)$

$= {\left(4 - 2 w - 2 {w}^{2} + {w}^{3}\right)}^{2}$

$= {\left(4 - 2 \left(w + {w}^{2}\right) + 1\right)}^{2}$

$= {\left(4 - 2 \left(\left(- 1\right)\right) + 1\right)}^{2}$

$= {7}^{2} = 49$