JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \( x^{2}+x+1=0 \) then the value of \( (x+\frac{1}{x})^{4}+(x^{2}+\frac{1}{x^{2}})^{4}+(x^{3}+\frac{1}{x^{3}})^{4}+...+(x^{25}+\frac{1}{x^{25}})^{4} \) is:
- A 128
- B 162
- C 175
- D 145
Answer & Solution
Correct Answer
(D) 145
Step-by-step Solution
Detailed explanation
\(x^2+x+1=0\) \(\Rightarrow x=\omega\) or \(\omega^2\) \(\therefore \alpha=\omega \& \beta=\omega^2\) \(=\left(\omega+\omega^2\right)^4+\left(\omega^2+\omega^4\right)^4+\left(\omega^3+\omega^6\right)^4+\ldots+\)\(\left(\omega^{25}+\omega^{50}\right)^4\)…
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