<s>[INST] 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.
 [/INST] 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., $xyz=-1$.
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