Question

w is a complex cube root unity than show that (1-w)(1-w²)(1-w⁴)(1-w⁵)=9​

Answer 1

Step-by-step explanation:

We have,

$(1 - w)(1 - {w}^{2} )(1 - {w}^{4} )(1 - {w}^{5} )$

$= (1 - w)(1 - {w}^{2} )(1 - {w}^{3}.w )(1 - {w}^{3}. {w}^{2} )$

$= (1 - w)(1 - {w}^{2} )(1 - w )(1 - {w}^{2} )$

$= \{ (1 - w)(1 - {w}^{2} ) \} ^{2}$

$= \{ 1 - w - {w}^{2} - {w}^{3} \} ^{2}$

$= ( 1 - w - {w}^{2} - 1 ) ^{2}$

$= ( - w - {w}^{2} ) ^{2}$

$= ( w + {w}^{2} ) ^{2}$

$= ( w)^{2} + {(w)}^{4} + 2 { w}^{3}$

$= ( w)^{2} + {(w)}^{3}.w + 2 { w}^{3}$

$= w^{2} + w + 2$

$= \frac{ - 1 - i \sqrt{3} }{2} + \frac{ - 1 + i \sqrt{3} }{2} + 2$

$= \frac{ - 1 - i \sqrt{3} - 1 + i \sqrt{3} + 4 }{2} +$

$= \frac{ - 2 + 4 }{2}$

$= \frac{ 2}{2} = 1$