Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$ Solve for $x$ in
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?
The equation I got was:
$$x^4-3x^3-22x^2-48x-32=0$$

Expand $$x^4-3 x^3-22 x^2-48 x-32=0$$
Try to factorize $$x^4-3 x^3-22 x^2-48 x-32=\left(a x+b+x^2\right) \left(c x+d+x^2\right)$$
RHS gives $$x^4+x^3 (a+c)+x^2 (a c+b+d)+x (a d+b c)+b d$$
thus we must have
$$
\begin{cases}
a+c=-3\\
ac+b+d=-22\\
ad+bc=-48\\
bd=-32\\
\end{cases}
$$
Final factorization should be
$$\left(x^2-6 x-8\right) \left(x^2+3 x+4\right)=0$$
