$$x^3+\frac{1}{x^3} = 0$$
Answer
$$ \begin{matrix}x_1 = 0.86603+0.5i & x_2 = 0.86603-0.5i & x_3 = -0.86603+0.5i \\[1 em] x_4 = -0.86603-0.5i & x_5 = 1i & x_6 = -i \end{matrix} $$
Explanation
$$ \begin{aligned} x^3+\frac{1}{x^3} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3 }. \\[1 em]x^3x^3+x^3\cdot\frac{1}{x^3} &= x^3\cdot0&& \text{cancel out the denominators} \\[1 em]x^6+1 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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