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TS EAMCET · Maths · Complex Number

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

  1. A \(\alpha^2\)
  2. B 0
  3. C \(-\alpha^2\)
  4. D \(\alpha\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(-\alpha^2\)

Step-by-step Solution

Detailed explanation

Given that \(x^6=1 \Rightarrow x^6-1=0\) ...(i) \(\begin{aligned} & \Rightarrow(x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0 \\ & \Rightarrow \quad x^5+x^4+x^3+x^2+x+1=0 \\ & \end{aligned}\) \([\because\) roots are non-real] Since \(\alpha\) is a root of the equation (i)…