Answer
$(z-y)(y-x)(z-x)$
Work Step by Step
The determinant of a $2 \times 2$ matrix can be computed by using the formula $det =ps-qr$
where $det =\begin{vmatrix}p & q \\r & s\end{vmatrix}$
We have $\begin{vmatrix}1 & x &x^2 \\1 & y & y^2 \\1 & z & z^2\end{vmatrix}=1\begin{vmatrix}y & y^2 \\z & z^2\end{vmatrix}-1 \begin{vmatrix}x & x^2 \\z & z^2\end{vmatrix}+1 \begin{vmatrix}x & x^2 \\y & y^2\end{vmatrix}$
or, $=yz^2-zy^2-xz^2+zx^2+xy^2-yx^2$
or, $=yz^2-zy^2-xz^2+zx^2+xy^2-yx^2+xyz-xyz$
or, $=z(yz+x^2-xz-xy)-y(yz+x^2-xz-xy)$
or, $=(z-y)(y-x)(z-x)$
