$p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$ $p$,$q$ and $r$ are roots of $x^3+2x^2+3x+3=0$. Find value of 
$$\sum_{cyc}\left(\frac{p}{p+1}\right)^3$$
I have modified the equations as  $$(x+1)^3=x^2-2$$ so 
$$ \frac{x^3}{(x+1)^3}=\frac{x^3}{x^2-2}=x+\frac{1}{x-\sqrt{2}}+\frac{1}{x+\sqrt{2}}$$ So
$$\sum_{cyc}\left(\frac{p}{p+1}\right)^3=\sum_{cyc}p+\sum_{cyc}\frac{1}{p-\sqrt{2}}+\sum_{cyc}\frac{1}{p+\sqrt{2}}$$
After this i have find equation whose roots are $p-\sqrt{2}$, $q-\sqrt{2}$ and $r-\sqrt{2}$ and similarly for other cyclic expression. but its too lengthy

There is an easier way if $y=\frac{1}{x}$ then $y$ is a root of 
$$3y^3+3y^2+2y+1=0$$ if $z=y+1$ then $z$ is a root of 
$$3z^3-6z^2-6z-2=0$$ and if $w=\frac{1}{z}$ then $w$ is a root of 
$$2w^3+6w^2+6w-3=0$$ and in the highly unlikely event I have not made a mistake, you can use Newton formula to find the sum of the cubes of the roots of this last equation.
