Question
$$(18x^3+27x^6+36x^5)\cdot9x^3 = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} (18x^3+27x^6+36x^5)\cdot9x^3 &= 0&& \text{simplify left side} \\[1 em](162x^3+243x^6+324x^5)x^3 &= 0&& \\[1 em]162x^6+243x^9+324x^8 &= 0&& \\[1 em]243x^9+324x^8+162x^6 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 243x^{9}+324x^{8}+162x^{6} = 0 } $, first we need to factor our $ x^6 $.
$$ 243x^{9}+324x^{8}+162x^{6} = x^6 \left( 243x^{3}+324x^{2}+162 \right) $$$ x = 0 $ is a root of multiplicity $ 6 $.
The remaining roots can be found by solving equation $ 243x^{3}+324x^{2}+162 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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