If α is a non real root of x6=1 then α5+α3+α+1α2+1=

is a root of x6=1 ⇒ α6=1, 1+α+α2+α3+α4+α5=0∴ α5+α3+α+1=-α2-α4α5+α3+α+1=-α21+α2 ⇒ α5+α3+α+11+α2=-α2

If α is a non real root of x6=1 then α5+α3+α+1α2+1=

is a root of x6=1 ⇒ α6=1, 1+α+α2+α3+α4+α5=0∴ α5+α3+α+1=-α2-α4α5+α3+α+1=-α21+α2 ⇒ α5+α3+α+11+α2=-α2

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

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$α^{2}$

$- α^{2}$

$α$

answer is B.

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Detailed Solution

is a root of $x^{6} = 1$

$\begin{array}{rcl} \Rightarrow & α^{6} = 1, 1 + α + α^{2} + α^{3} + α^{4} + α^{5} = 0 \\ ∴ α^{5} + α^{3} + α + 1 & = & - α^{2} - α^{4} \\ α^{5} + α^{3} + α + 1 & = & - α^{2} (1 + α^{2}) \\ \Rightarrow & \frac{α^{5} + α^{3} + α + 1}{1 + α^{2}} = - α^{2} \end{array}$