If \(x + \frac{1}{x} = -2\), find the value of \(x^8 + \frac{1}{x^8} + 2x^5 + \frac{2}{x^5} + x^2 + \frac{1}{x^2}\).
0
2
-2
4
Correct answer: 0
Solution
From \(x + \frac{1}{x} = -2\), we get \(x^2 + 2x + 1 = 0\), so \((x+1)^2=0\) and hence \(x=-1\). Substituting \(x=-1\) into the expression gives cancellation of all terms, so the value is 0.