<s>[INST] 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:
$$x^4-3x^3-22x^2-48x-32=0$$
 [/INST] Note that the given equation
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
is purposefully structured to allow convenient factorization as follows
$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$
$$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$
$$x^4-(3x+4)x^2-2(3x+4)^2 =0$$
$$(x^2+(3x+4))(x^2-2(3x+4))=0$$
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