$$\frac{x^3}{6}+\frac{1}{2}x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \sqrt{ 3 } i & x_3 = -\sqrt{ 3 } i \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x^3}{6}+\frac{1}{2}x &= 0&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{x^3}{6}+6\frac{1}{2}x &= 6\cdot0&& \text{cancel out the denominators} \\[1 em]x^3+3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}+3x = 0 } $, first we need to factor our $ x $.
$$ x^{3}+3x = x \left( x^{2}+3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{2}+3 = 0$.
$ x^{2}+3 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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