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

$\bf{I\; have \;Solved\; Like \;This \; Way}$ 
$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0\Rightarrow 512x^9+(9\cdot 64)x^6+(27\cdot 8)x^3-512x+219=0$$
So $$\underbrace{(8x^3)^3+3(8x^3)^2\cdot 3+3(3^2)\cdot 8x^3+3^3}-512x+219-3^3=0$$
So $$(8x^3+3)^3=512x-219\Rightarrow (8x^3+3)^3=512\left(x-\frac{192}{512}\right)=8^3\left(x-\frac{3}{8}\right)^{\frac{1}{3}}.$$
So $$8^3\left(x^3+\frac{3}{8}\right) = 8\left(x-\frac{3}{8}\right)^{\frac{1}{3}}\Rightarrow x^3+\frac{3}{8} = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}$$
Now Let $\displaystyle f(x)=x^3+\frac{3}{8},$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;,$ Then $\displaystyle f^{-1}(x) = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}\;,$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;$
So We have To solve $$f(x) = f^{-1}(x)$$
Now We now that $f(x)$ and $f^{-1}(x)$ is Symmetrical about $y=x$ line.
So $$f(x) = f^{-1}(x) =x$$
So $$x^3+\frac{3}{8}=x\Rightarrow 8x^3-8x+3=0$$
So $$(2x-1)\left[4x^2+2x-3\right]=0\Rightarrow x=\frac{1}{2}\;\;,x=\frac{-1\pm \sqrt{13}}{4}$$
