If - is a non real root of the equation x6 - 1 - 0 then -2 - -3 - -4 - -5- - 1 -

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Relations between Roots and Coefficients : Higher Order Equations

If α is a n...

Question

If $α$ is a non real root of the equation $x^{6} - 1 = 0$ then $\frac{α^{2} + α^{3} + α^{4} + α^{5}}{α + 1} =$

α

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Solution

The correct option is D $- 1$
Sum and product of roots of $x^{6} - 1 = 0$ are
$α^{6} = 1$

$1 + α + α^{2} + α^{3} + α^{4} + α^{5} = 0$

$α^{2} + α^{3} + α^{4} + α^{5} = - (1 + α)$
Given $= \frac{α^{2} + α^{3} + α^{4} + α^{5}}{1 + α}$
$= - 1$

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If α is a non real root of the equation x6−1=0 then α2+α3+α4+α5α+1=

The correct option is D −1Sum and product of roots of x6−1=0 areα6=11+α+α2+α3+α4+α5=0α2+α3+α4+α5=−(1+α) Given =α2+α3+α4+α51+α =−1

If $α$ is a non real root of the equation $x^{6} - 1 = 0$ then $\frac{α^{2} + α^{3} + α^{4} + α^{5}}{α + 1} =$

The correct option is D $- 1$
Sum and product of roots of $x^{6} - 1 = 0$ are
$α^{6} = 1$

$1 + α + α^{2} + α^{3} + α^{4} + α^{5} = 0$

$α^{2} + α^{3} + α^{4} + α^{5} = - (1 + α)$
Given $= \frac{α^{2} + α^{3} + α^{4} + α^{5}}{1 + α}$
$= - 1$