Find the value of: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$ 
Let $a,b,c$ be roots of the cubic
$$x^3-x^2-2x+1=0$$
Then, find the value of:
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}$$

My attempt.
I used the substitutions $$a+b=x^3, b+c=y^3, a+c=z^3$$
$$x^3+y^3+z^3=2(a+b+c)=2$$
Then I used the identity $$x^3+y^3+z^3=(x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz$$
But I stuck here. I can not simplify $$xy+yz+xz$$
 A: From the given cubic, we know that $a+b+c=1$, therefore
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=\sqrt[3]{1-c}+\sqrt[3]{1-a}+\sqrt[3]{1-b}=\sqrt[3]{A}+\sqrt[3]{B}+\sqrt[3]{C}$$
where $A,B,C$ are the roots of
$(1-x)^3-(1-x)^2-2(1-x)+1=0$, that is $x^3-2 x^2-x+1=0$.
It can be shown (see for example this link) that
$$x^3+(\alpha^3−3\alpha\beta +3\gamma)x^2+
(\beta^3−3\alpha\beta \gamma+3\gamma^2)x+\gamma^3=0$$
is the cubic equation whose roots are the cubes of the roots of $x^3+\alpha x^2+\beta x+\gamma=0$. So it remains to find $-\alpha$ in the system
$$\begin{cases}
\alpha^3−3\alpha\beta +3\gamma=-2\\
\beta^3−3\alpha\beta \gamma+3\gamma^2=-1\\
\gamma^3=1
\end{cases}$$
which boils down to $\gamma=1$, $\beta=\frac{\alpha^3+5}{3\alpha}$ and
$$(\alpha^3+5)^3-27\alpha^3(\alpha^3+5)+4\cdot 27\alpha^3=0$$
or
$$\alpha^9-12\alpha^6+48\alpha^3+125=(\alpha^3-4)^3+189=0$$
which has just one real root at $\alpha=-\sqrt[3]{3\sqrt[3]{7}-4}$.
Hence
$$\sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{a+c}=-\alpha=\sqrt[3]{3\sqrt[3]{7}-4}\approx 1.20249.$$
P.S. Notice that the roots $a,b,c$ of the cubic polynomial $f(x)=x^3-x^2-2x+1$ are three distinct real numbers (one is negative and two are positive) because $f(−2)<0$, $f(0)>0$, $f(1)<0$, $f(2)>0$.
