+535 Let a be a real number such that a + \frac{1}{a} + a^2 + \frac{1}{a^2} = 0. Compute a^5.
+9673 Multiplying all terms by a^2 gives
\(a^4 + a^3 + a + 1 = 0\\ (a^3 + 1)(a + 1) = 0\\ (a + 1)^2(a^2 - a + 1) = 0\\ a = -1\text{ or }a^2 - a + 1 = 0\)
Note that the equation \(a^2 - a +1 = 0\) has determinant \(\Delta = (-1)^2 - 4(1)(1) = -3 < 0\), so \(a^2 - a +1 = 0\) has no real solutions.
Hence, a = -1 is the only possibility. Then \(a^5 = (-1)^5 = -1\).
