$$\frac{x^5-x^3+x-1}{x^3+x+1} = 0$$
Answer
$$ \begin{matrix}x_1 = 1 & x_2 = 0.51891+0.66661i & x_3 = 0.51891-0.66661i \\[1 em] x_4 = -1.01891+0.60257i & x_5 = -1.01891-0.60257i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x^5-x^3+x-1}{x^3+x+1} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3+x+1 }. \\[1 em](x^3+x+1)\frac{x^5-x^3+x-1}{x^3+x+1} &= (x^3+x+1)\cdot0&& \text{cancel out the denominators} \\[1 em]x^5-x^3+x-1 &= 0&& \\[1 em] \end{aligned} $$
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