Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$ 
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$

We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me


Using factor theorem:
If $x-a$ divides $f(x)$, then $f(a)=0$.

As $(\frac{1}{2})^{9}=\frac{1}{512}$, it's worth to try $x=\pm \frac{1}{2}$.
Substituting $x=\frac{1}{2}$, LHS vanishes, so $\frac{1}{2}$ is one possible value.
Further factorize gets,
$$\left( x-\frac{1}{2} \right)
 \left( x^{2}-\frac{x}{2}-\frac{3}{4} \right)
 \left( x^{6}+x^{4}+\frac{3x^{3}}{4}+x^{2}+\frac{3x}{8}+\frac{7}{3} \right)=0$$ and the second factor can be solved by quadratic formula.
