<s>[INST] Real roots of an equation A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that madness calculation. I reduced it to $$x^3+1=2(2x-1)^{\frac{1}{3}}$$ I don't have any  useful attempts.
EDIT: After several tries I think I found something useful. After getting to this form $x^3-2(2x-1)^{\frac{1}{3}}+1=0$ we might notice that it's similar to the equation that you provided me in the answers namely$$(x-1)(x^2 +x-1)=x^3 - 2x+1=0$$ Now in order that this to give me the solution, we must have that $-2(2x-1)^{\frac{1}{3}}=-2x\, $ If we cube both sides we return to $x^3 - 2x+1=0$ (the solutions) Is this a repeatedly loop in the equation? Can this be useful? 
 [/INST] The polynomial can be factored by considering the product of 
$$(x^3 + a x^2 + b x + c) \cdot (x^6 + \alpha x^5 + \beta x^4 + \gamma x^3 + \delta x^2 + \eta x + \mu).$$
With a little bit of effort it can be found that 
$$(x - 1) (x^2 + x - 1) (x^6 + 2 x^4 + 2 x^3 + 4 x^2 + 2 x + 9) = 0.$$
The real roots are then obtained as $x=1$ and roots from $x^2 + x -1=0$.
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