Question

If \( x^2 + x + 1 = 0 \), then \[ \left( x + \frac{1}{x} \right)^4 + \left( x^2 + \frac{1}{x^2} \right)^4 + \left( x^3 + \frac{1}{x^3} \right)^4 + \dots + \left( x^{25} + \frac{1}{x^{25}} \right)^4 \] is

Hint:

For problems involving cube roots of unity, recognize the repeating patterns and simplify the powers of \( x \) accordingly.

Answer 1

Correct Answer: 145

The equation \(x^2 + x + 1 = 0\) has roots \(x = e^{2\pi i/3}\) and \(x^2 = \frac{-1-x}{x}\). To solve this, rewrite: For \(x + \frac{1}{x} = -1\): Now compute powers: 1) \(x^2 + \frac{1}{x^2} = (-1)^2 - 2 \cdot 1 = 1\) 2) \(x^3 + \frac{1}{x^3} = x^3 + (-1/x^3) = 0\) as \(x^3 = -1\) so \(-x^3 = 1\). Notice a recurring sequence: For any \(n\), \(x^n + \frac{1}{x^n} = -1\text{ if }n \equiv 1 \pmod{3}\), \(1 \text{ if }n \equiv 2 \pmod{3}\), \(0\text{ if }n \equiv 0 \pmod{3}\). The problem requires: \(\left( x^n + \frac{1}{x^n} \right)^4\) summed over terms 1 to 25. Dividing the sequence from 1 to 25 modulo 3 gives:\ Terms 1, 4, 7... (1 modulo 3): There are 9 such terms.\ Terms 2, 5, 8... (2 modulo 3): There are 9 such terms. Terms 3, 6, 9... (0 modulo 3): There are 8 such terms (contribute 0 since \(\left(0\right)^4 = 0\)). \(\left(x^n + \frac{1}{x^n} \right)^4\) for these cases: 1. \((-1)^4 = 1\) 2. \(1^4 = 1\) Sum over required terms gives \(9\cdot 1 + 9 \cdot 1 = 18\). Verify range: \(18\) fits within 145,145.