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MathsJEE Main 21 Jan 2026, Shift 1·Q17·MCQ·Moderate

Complex Numbers · Cube roots of unity

If x2+x+1=0x^2 + x + 1 = 0, then the value of (x+1x)4+(x2+1x2)4+(x3+1x3)4+…+(x25+1x25)4\left(x + \frac1x\right)^4 + \left(x^2 + \frac{1}{x^2}\right)^4 + \left(x^3 + \frac{1}{x^3}\right)^4 + \ldots + \left(x^{25} + \frac{1}{x^{25}}\right)^4 is:
  1. (1)128
  2. (2)162
  3. (3)175
  4. (4)145

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