In a determinant prove $xyz = -1$

Question

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.

Answer 1

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 $xyz=-1$.

Answer 2

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., $xyz=-1$.

Answer 3

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 $R_1'=R_1-x\cdot R_2$ $$\begin{vmatrix} x^3+1 & x^2 & x \\ y^2+xy+x^2 & y+x &1 \\ z+y+x & 1 &0 \\ \end{vmatrix}=\begin{vmatrix} 1-xy(x+y) & -xy &0 \\ y^2+xy+x^2 & y+x &1 \\ z+y+x & 1 &0 \\ \end{vmatrix}=?$$