Question
$$(x^2+3)x(x^4+9-3x^2) = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} (x^2+3)x(x^4+9-3x^2) &= 0&& \text{simplify left side} \\[1 em](x^3+3x)(x^4+9-3x^2) &= 0&& \\[1 em]x^7+9x^3-3x^5+3x^5+27x-9x^3 &= 0&& \\[1 em]x^7+9x^3-3x^5+3x^5+27x-9x^3 &= 0&& \\[1 em]x^7+27x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{7}+27x = 0 } $, first we need to factor our $ x $.
$$ x^{7}+27x = x \left( x^{6}+27 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{6}+27 = 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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