$$\frac{1}{3x^2} = \frac{x+3}{2x^2}-\frac{1}{6x^2}$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -2 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{3x^2} &= \frac{x+3}{2x^2}-\frac{1}{6x^2}&& \text{multiply ALL terms by } \color{blue}{ 3x^2\cdot2\cdot6 }. \\[1 em]3x^2\cdot2\cdot6\cdot\frac{1}{3x^2} &= 3x^2\cdot2\cdot6 \cdot \frac{x+3}{2x^2}-3x^2\cdot2\cdot6\cdot\frac{1}{6x^2}&& \text{cancel out the denominators} \\[1 em]12x^4 &= 18x^5+54x^4-6x^4&& \text{simplify right side} \\[1 em]12x^4 &= 18x^5+48x^4&& \text{move all terms to the left hand side } \\[1 em]12x^4-18x^5-48x^4 &= 0&& \text{simplify left side} \\[1 em]-18x^5-36x^4 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -18x^{5}-36x^{4} = 0 } $, first we need to factor our $ x^4 $.
$$ -18x^{5}-36x^{4} = x^4 \left( -18x-36 \right) $$$ x = 0 $ is a root of multiplicity $ 4 $.
The second root can be found by solving equation $ -18x-36 = 0$.
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