Polynomial Roots - 3

We can immediately spot lots of powers of $3$ in the coefficients. Also, some coefficients involve a $\sqrt6$ , and some don't.

Rearranging, the equation is $\left(243x^6-18x^2-3 \right) \sqrt6 = -2x \left(27x^4+9x ^2+1 \right)$

Now, let $u=3x$ : $\left(\frac13 u^6-2u^2-3 \right) \sqrt6 = -\frac23 u \left(\frac13 u^4+u ^2+1 \right)$

Clearing fractions: $3\left(u^6-6u^2-9 \right) \sqrt6 = -2u \left(u^4+3u ^2+3 \right)$

Noting that $u^2-3$ is a factor of the left-hand side, $3\left(u^2-3\right) \left(u^4+3u^2+3 \right) \sqrt6 = -2u \left(u^4+3u ^2+3 \right)$

For real $u$ , the expression $u^4+3u ^2+3$ is always positive; hence we can divide through leaving $3\left(u^2-3\right) \sqrt6 = -2u$

This is just a quadratic in $u$ : $3\sqrt6 u^2 + 2u - 9\sqrt6=0$

Solving, $u=\frac{-2\pm \sqrt{4+648}}{6\sqrt6}=\frac{-1\pm \sqrt{163}}{3\sqrt6}$

To get a positive result, we need to take the positive root; finally $x=\frac13 u = \frac{-1+ \sqrt{163}}{9\sqrt6}$

This can be rewritten as $x=\sqrt{\frac{163}{486}}-\sqrt{\frac{1}{486}}$

so the final answer is $\boxed{163}$ .