(Almost) degree 6 equation

Let $A = x^{2}+9$ and $B = 6x$ so that $A+B = (x+3)^{2}$

The equation becomes $A^{3} + B^{3} = (A+B)^{3}$ .

$3AB(A+B) = 0$

$A = 0 \text{ or } B = 0 \text{ or } A = -B$

$x^{2}+9 = 0$ gives you 2 complex roots.

$6x = 0$ gives $x = 0$ .

$x^{2} + 9 = -6x$ gives $x = -3$ .

Therefore $\alpha^{6} + \beta^{6} = \boxed{729}$ ~~~

Well I spotted something here:

You can see $x^2+9=\left(x+3\right)^2-6x$ . So using this we get

$\displaystyle\left(\left(x+3\right)^2-6x\right)^3+216x^3=\left(x+3\right)^6$

$\rightarrow\,\left(x+3\right)^6-216x^3-18x\left(x+3\right)^2\left(x^2+9\right)^2+216x^3=\left(x+3\right)^6$

$\rightarrow\,x\left(x+3\right)^2\left(x^2+9\right)^2=0$

Now since $x\in\mathbb{R}$ so $x^2+9\geq\,9$ . Thus this leads down to $x=0,-3$ . So $\left(\alpha,\beta\right)=$ some permutation of $\left(0,-3\right)$ . So we have $\alpha^6+\beta^6=\boxed{729}$ .