$$\frac{27x^5+9x^4-18x^3}{9}x^2 = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1 & x_3 = \dfrac{ 2 }{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{27x^5+9x^4-18x^3}{9}x^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 9 }. \\[1 em]9 \cdot \frac{27x^5+9x^4-18x^3}{9}x^2 &= 9\cdot0&& \text{cancel out the denominators} \\[1 em]27x^7+9x^6-18x^5 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 27x^{7}+9x^{6}-18x^{5} = 0 } $, first we need to factor our $ x^5 $.
$$ 27x^{7}+9x^{6}-18x^{5} = x^5 \left( 27x^{2}+9x-18 \right) $$$ x = 0 $ is a root of multiplicity $ 5 $.
The remaining roots can be found by solving equation $ 27x^{2}+9x-18 = 0$.
$ 27x^{2}+9x-18 = 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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