Question
$$(x^2+3x-\frac{18}{x^2}-36)(x-\frac{6}{x}+4) = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} (x^2+3x-\frac{18}{x^2}-36)(x-\frac{6}{x}+4) &= 0&& \text{simplify left side} \\[1 em](\frac{x^4+3x^3-18}{x^2}-36)(\frac{x^2-6}{x}+4) &= 0&& \\[1 em]\frac{x^4+3x^3-36x^2-18}{x^2}\frac{x^2+4x-6}{x} &= 0&& \\[1 em]\frac{x^6+7x^5-30x^4-162x^3+198x^2-72x+108}{x^3} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3 }. \\[1 em]x^3\frac{x^6+7x^5-30x^4-162x^3+198x^2-72x+108}{x^3} &= x^3\cdot0&& \text{cancel out the denominators} \\[1 em]x^{12}+7x^{11}-30x^{10}-162x^9+198x^8-72x^7+108x^6 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{12}+7x^{11}-30x^{10}-162x^{9}+198x^{8}-72x^{7}+108x^{6} = 0 } $, first we need to factor our $ x^6 $.
$$ x^{12}+7x^{11}-30x^{10}-162x^{9}+198x^{8}-72x^{7}+108x^{6} = x^6 \left( x^{6}+7x^{5}-30x^{4}-162x^{3}+198x^{2}-72x+108 \right) $$$ x = 0 $ is a root of multiplicity $ 6 $.
The remaining roots can be found by solving equation $ x^{6}+7x^{5}-30x^{4}-162x^{3}+198x^{2}-72x+108 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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