Answer
See below
Work Step by Step
We have $\det(A)=\begin{vmatrix}
1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2
\end{vmatrix}\\
=\begin{vmatrix}
1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 1 & z-x & z^2-x^2
\end{vmatrix}$
Using P2:
$\det(A)=(y-x)(z-x)\begin{vmatrix}
1 & x & x^2 \\ 0 & 1& y+x \\ 0& 1 & z+x
\end{vmatrix}\\
=(y-x)(z-x)\begin{vmatrix}
1 & x & x^2 \\ 0 & 1& y+x \\ 0& 0 & z-y
\end{vmatrix}$
Hence, according to Theorem 3.2.1, we have $\det(A)=(y-x)(z-x)(z-y)\\=(y-z)(z-x)(x-y)$
