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.
 A: This is my answer (for sure it isn't the best approach). Applying cofactors we have
$$(x^3+1)\begin{vmatrix}y^2 & y\\z^2 & z\end{vmatrix}-x^2\begin{vmatrix}y^3+1 & y\\z^3+1 & z\end{vmatrix}+x\begin{vmatrix}y^3+1 & y^2\\z^3+1 & z^2\end{vmatrix}=0.$$
After some operations we get $$(x^3+1)yz(y-z)-x^2(y-z)(yz(y+z)-1)+x(y-z)(y^2z^2-y-z)=0.$$ Then $$(y-z)(x^3yz+yz-x^2y^2z-x^2yz^2+x^2+xy^2z^2-xy-xz)=0.$$
Since $y\neq z$, we deduce that $x^3yz+yz-x^2y^2z-x^2yz^2+x^2+xy^2z^2-xy-xz=0$. Finally we factorize the LHS and we get $$xyz(x^2-xy-xz+yz)+(x^2+yz-xy-xz)=(x^2-xy-xz+yz)(xyz+1)=$$ $$=(x-z)(y-z)(xyz+1)=0.$$
But $x\neq y$ and $x\neq z$, hence it must be $xyz+1=0$, i.e., $xyz=-1$.
A: 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 $R_1'=R_1-x\cdot R_2$  $$\begin{vmatrix}
        x^3+1 & x^2 & x \\
        y^2+xy+x^2 & y+x &1 \\
        z+y+x & 1 &0 \\
        \end{vmatrix}=\begin{vmatrix}
        1-xy(x+y) & -xy &0 \\
        y^2+xy+x^2 & y+x &1 \\
        z+y+x & 1 &0 \\
        \end{vmatrix}=?$$
A: Outline:
(1) Break up the determinant as a sum of two determinants based on splitting
out the $+1$ terms.
(2) The given determinant can now be expressed as $xyzD + D = D(xyz + 1)$.
(3) Show that $D=-(x - y)(y - z)(z - x)$ [look up Vandermonde matrix]
(4) Since $x,y,z$ are distinct, $D$ is nonzero.
(5) Since the given determinant is $0$, it follows that $xyz=-1$.
