$$x^2+\frac{1}{x^2}+x^6+\frac{1}{x^6} = 0$$
Answer
$$ \begin{matrix}x_1 = 0.70711+0.70711i & x_2 = 0.70711-0.70711i & x_3 = -0.70711+0.70711i \\[1 em] x_4 = -0.70711-0.70711i & x_5 = 0.25882+0.96593i & x_6 = 0.25882-0.96593i \\[1 em] x_7 = -0.96593+0.25882i & x_8 = -0.96593-0.25882i & x_9 = -0.25882+0.96593i \\[1 em] x_10 = -0.25882-0.96593i & x_11 = 0.96593+0.25882i & x_12 = 0.96593-0.25882i \end{matrix} $$
Explanation
$$ \begin{aligned} x^2+\frac{1}{x^2}+x^6+\frac{1}{x^6} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2x^6 }. \\[1 em]x^2x^6x^2+x^2x^6\cdot\frac{1}{x^2}+x^2x^6x^6+x^2x^6\cdot\frac{1}{x^6} &= x^2x^6\cdot0&& \text{cancel out the denominators} \\[1 em]x^{10}+x^2+x^{14}+\frac{1}{x^2} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2 }. \\[1 em]x^2\cdot1x^{10}+x^2\cdot1x^2+x^2\cdot1x^{14}+x^2\cdot\frac{1}{x^2} &= x^2\cdot0&& \text{cancel out the denominators} \\[1 em]x^{12}+x^4+x^{16}+1 &= 0&& \text{simplify left side} \\[1 em]x^{16}+x^{12}+x^4+1 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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