Question

If 1,$\alpha$,${\alpha }^2$,${\alpha }^3$,.................${\alpha }^6$ are the roots of the equation x = $(-1{)}^{\frac{1}{7}}$.Find the value of

${\alpha }^4$ + ${\alpha }^6$ + ${\alpha }^7$ + ${\alpha }^5$ + ${\alpha }^7$ + ${\alpha }^8$ + ${\alpha }^7$ + ${\alpha }^9$ + ${\alpha }^{10}$

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Answer 1

x = $(1{)}^{\frac{1}{7}}$

We Know, sum of the roots

( 1 + $\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$) = 0

and ${\alpha }^7$ = 1

Where ever we see that $\alpha$ power more than 7.Replace $a^7$ = 1

For eg. ${\alpha }^8$ = ${\alpha }^7$ $\times$ $\alpha$ = $\alpha$,

${\alpha }^9$ = ${\alpha }^2$, ${\alpha }^{10}$ = ${\alpha }^3$

So, ${\alpha }^4$ + ${\alpha }^6$ + ${\alpha }^7$ + ${\alpha }^5$ + ${\alpha }^7$ + ${\alpha }^8$ + ${\alpha }^7$ + ${\alpha }^9$ + ${\alpha }^{10}$

= ${\alpha }^4$ + ${\alpha }^6$ + 1 + ${\alpha }^5$ + 1 + $\alpha$ + 1 +${\alpha }^2$ + ${\alpha }^3$

= (1 + $\alpha$ + ${\alpha }^2$ + ${\alpha }^3$ + ${\alpha }^4$ + ${\alpha }^5$ + ${\alpha }^6$) + 2

We know sum of all the roots of unity = 0

= 0 + 2 = 2