TS EAMCET · Maths · Complex Number
If \(\alpha\) is a root of the equation \(x^2-x+1=0\) then \(\left(\alpha+\frac{1}{\alpha}\right)^3+\left(\alpha^2+\frac{1}{\alpha^2}\right)^3+\left(\alpha^3+\frac{1}{\alpha^3}\right)^3+\left(\alpha^4+\frac{1}{\alpha^4}\right)^3+\ldots\) to 12 terms \(=\)
- A \(-32\)
- B \(32\)
- C \(0\)
- D \(16\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
\(\alpha^2-\alpha+1=0 \implies \alpha+\frac{1}{\alpha}=1 \) \(\alpha^2-\alpha+1=0 \implies (\alpha+1)(\alpha^2-\alpha+1)=0 \implies \alpha^3+1=0 \implies \alpha^3=-1 \) \(\left(\alpha+\frac{1}{\alpha}\right)^3 = (1)^3 = 1 \)…
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