TS EAMCET · Maths · Complex Number
If \(\alpha\) is a non-real root of the equation \(x^6-1=0\), then \(\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1}\) is equal to
- A \(\alpha\)
- B \(1\)
- C \(0\)
- D \(-1\)
Answer & Solution
Correct Answer
(D) \(-1\)
Step-by-step Solution
Detailed explanation
Since, \(\alpha\) is a non-real root of the equation \(x^6-1=0\).…
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