Question
$$x+\frac{1}{x^3}+4x^2+6x+4 = 0$$
Answer
$$ x = -1 $$
Explanation
$$ \begin{aligned} x+\frac{1}{x^3}+4x^2+6x+4 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3 }. \\[1 em]x^3x+x^3\cdot\frac{1}{x^3}+x^3\cdot4x^2+x^3\cdot6x+x^3\cdot4 &= x^3\cdot0&& \text{cancel out the denominators} \\[1 em]x^4+1+4x^5+6x^4+4x^3 &= 0&& \text{simplify left side} \\[1 em]4x^5+7x^4+4x^3+1 &= 0&& \\[1 em] \end{aligned} $$
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