In a determinant prove $xyz = -1$ If we are given the following determinant
$$\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^3+1 & y^2 & y \\
        z^3+1 & z^2 & z \\
        \end{vmatrix}=0 
$$
and $x, y, z$ are all different, then we have to prove that $xyz = -1$.
I tried to expand the determinant, but using that, it is getting too complicated.

Hint:
Apply $R'_2=R_2-R_1$  and  $R'_3=R_3-R_1$ to find 
$$\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^3+1 & y^2 & y \\
        z^3+1 & z^2 & z \\
        \end{vmatrix}
=\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^3-x^3 & y^2-x^2 & y-x \\
        z^3-x^3 & z^2-x^2 & z-x \\
        \end{vmatrix}$$  $$=(y-x)(z-x)\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z^2+zx+x^2 & z+x &1 \\
        \end{vmatrix}$$
Now apply $R_3'=R_3-R_2$  $$\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z^2+zx+x^2 & z+x &1 \\
        \end{vmatrix}=\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z^2-y^2+zx-xy & z-y &0 \\
        \end{vmatrix}=(z-y)\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z+y+x & 1 &0 \\
        \end{vmatrix}$$
Finally $R_1'=R_1-x\cdot R_2$  $$\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z+y+x & 1 &0 \\
        \end{vmatrix}=\begin{vmatrix}
        1-xy(x+y) & -xy &0 \\
        y^2+xy+x^2 & y+x &1 \\
        z+y+x & 1 &0 \\
        \end{vmatrix}=?$$
