How many solutions the following equation has?

$3x^3=\left(x^2+\sqrt{18}x+\sqrt{32}\right)\left(x^2-\sqrt{18}x-\sqrt{32}\right)-4x^2$

The given equation is,
$3{\mathrm{x}}^3=\left({\mathrm{x}}^2+\sqrt{18}\mathrm{x}+\sqrt{32}\right)\left({\mathrm{x}}^2-\sqrt{18}\mathrm{x}-\sqrt{32}\right)-4{\mathrm{x}}^2$

By solving the above equation, we get:

$\Rightarrow 3{\mathrm{x}}^3=\left({\mathrm{x}}^2+\sqrt{18}\mathrm{x}+\sqrt{32}\right)\left({\mathrm{x}}^2-\sqrt{18}\mathrm{x}-\sqrt{32}\right)-4{\mathrm{x}}^2$

[By using: $\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{=}\mathbf{ }\left(a^2-b^2\right)\mathbf{]}$

$\Rightarrow 3{\mathrm{x}}^3=\left({\mathrm{x}}^4-(\sqrt{18}\mathrm{x}+\sqrt{32}{)}^2\right)-4{\mathrm{x}}^2$       

[By using: ${\left(a+b\right)}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ab}\mathbf{]}$

$\Rightarrow 3{\mathrm{x}}^3=\left({\mathrm{x}}^4-\left(18{\mathrm{x}}^2+32+2\sqrt{18}\sqrt{32}\mathrm{x}\right)\right)-4{\mathrm{x}}^2$  

$\Rightarrow 3{\mathrm{x}}^3={\mathrm{x}}^4-18{\mathrm{x}}^2-32-2\sqrt{18}\sqrt{32}\mathrm{x}-4{\mathrm{x}}^2$ 

 $\Rightarrow {\mathrm{x}}^4-3{\mathrm{x}}^3-22{\mathrm{x}}^2-2\sqrt{18}\sqrt{32}\mathrm{x}-32=0$      

 $\mathbf{\Rightarrow }\mathbf{ }{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{22}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{48}\mathbf{x}\mathbf{-}\mathbf{32}\mathbf{=}\mathbf{0}$

As the highest power of the variable in the above equation is 4, its degree will be 4.

We know that the degree of a polynomial is equal to the number of solutions of the polynomial.

Thus, the $3{\mathrm{x}}^3=\left({\mathrm{x}}^2+\sqrt{18}\mathrm{x}+\sqrt{32}\right)\left({\mathrm{x}}^2-\sqrt{18}\mathrm{x}-\sqrt{32}\right)-4{\mathrm{x}}^2$ have 4 solutions.