Originally Posted by

**SlipEternal** Multiplying out: $\displaystyle x^2+x+2+\dfrac{1}{x} + \dfrac{1}{x^2} = 1$.

Multiply both sides by $\displaystyle x^2$ to get $\displaystyle x^4 + x^3 + x^2 +x + 1 = 0$.

Multiplying both sides by $\displaystyle x-1$ gives $\displaystyle x^5-1 = 0$ (but $\displaystyle x \neq 1$ since we added the root). Hence, $\displaystyle x = -(-1)^{1/5}, x= (-1)^{2/5}, x = -(-1)^{3/5}, x = (-1)^{4/5}$ are the four solutions. In each case, $\displaystyle x^5 = 1$. So:

$\displaystyle \left(x^{20} + \dfrac{1}{x^{20}} + 1\right)\left(x^{20} + \dfrac{1}{x^{20}}\right) = \left((x^5)^4 + \dfrac{1}{(x^5)^4} + 1\right)\left( (x^5)^4 + \dfrac{1}{(x^5)^4}\right) = 3\cdot 2 = 6$