Search any question & find its solution
Question:
Answered & Verified by Expert
If $\alpha$ is a non-real root of $x^6=1$, then $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}$ is equal to
Options:
Solution:
2736 Upvotes
Verified Answer
The correct answer is:
$-\alpha^2$
Given that
$\begin{aligned}
& \Rightarrow(x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0 \\
& \Rightarrow \quad x^5+x^4+x^3+x^2+x+1=0 \\
&
\end{aligned}$
$[\because$ roots are non-real]
Since $\alpha$ is a root of the equation (i)
$\begin{array}{ll}
\therefore & \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0 \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\left(\alpha^4+\alpha^2\right) \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\alpha^2\left(\alpha^2+1\right) \\
& \frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2
\end{array}$
$\begin{aligned}
& \Rightarrow(x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0 \\
& \Rightarrow \quad x^5+x^4+x^3+x^2+x+1=0 \\
&
\end{aligned}$
$[\because$ roots are non-real]
Since $\alpha$ is a root of the equation (i)
$\begin{array}{ll}
\therefore & \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0 \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\left(\alpha^4+\alpha^2\right) \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\alpha^2\left(\alpha^2+1\right) \\
& \frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.