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? 

Let $f(x)=(x^3+1)/2$. Then $f$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ with inverse $g(x)=(2x-1)^{1/3}$. As you already remarked, the original equation can be written as
$$((x^3+1)=2(2x-1)^{1/3}.$$
This equation is equivalent to $f(x)=g(x)$ or $f(f(x))=x.$
Now note that if $f(x)=x$, that is $x^3-2x+1=0$ then $f(f(x))=x$ which implies that $x^3-2x+1$ divides the given polynomial. After the division, we obtain the factorization
$$(x^3-2x+1)(\underbrace{(x^3+1)^2 + 2 x^4+3x^2+(x+1)^2}_{\geq 0} + 7) \tag{*}.$$
So it remains to solve 
$$x^3-2x+1=(x - 1) (x^2 + x - 1)=0$$ 
and the desired real roots are $x_1=1$, $x_2=(-1-\sqrt{5})/2$,  and $x_3=(-1+\sqrt{5})/2$ .
P.S. In my first answer I used maple to find the factorization $(*)$.
