# Solving for the roots of a sextic

I was solving the equation $3(1+x^2+x^3)^2=(2+x)^4$ for $x$, and after expanding it out, I got $$3x^6+6x^5+2x^4-2x^3-18x^2-32x-13=0\tag{1}$$ which should be solvable because it has a Galois group of order $72$. But since it's a degree six, I have no method for solving this.

I have attempted to factor it into two cubics, but the condition wasn't met, so I can't factor it. This polynomial has (maybe?) irrational roots so the rational root theorem won't work.

• Hint: write it as a difference of two squares, instead of expanding. – dxiv Oct 16 '16 at 16:51
• You can show that- a bit hard- the equation has no rational root. So think about the above comment. – Mikasa Oct 16 '16 at 16:55
• Is $$(\sqrt{3}(1+x^2+x^3)-(2+x)^2)(\sqrt{3}(1+x^2+x^3)+(2+x)^2)=0$$ – AsdrubalBeltran Oct 16 '16 at 17:15

From $3(1+x^2+x^3)^2=(2+x)^4$, we can move all terms to one side and use $a^2-b^2=(a+b)(a-b)$ to get\begin{align*} & (\sqrt3(1+x^2+x^3)-(2+x)^2)(\sqrt3(1+x^2+x^3)+(2+x)^2)=0\tag1\\ & (g\sqrt3+(\sqrt3-1)g^2-4g+\sqrt3-4)(g\sqrt3+(\sqrt3+1)g^2+4g+\sqrt3+4)=0\tag2\end{align*} And solving for the roots of the cubic gives all possible values of $g$.