Determinant of a $3 \times 3$ Vandermonde matrix

Question

Let $$A = \begin{bmatrix} p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{bmatrix}$$ Prove that $$\det(A) = (r-q)(r-p)(p-q)$$

Answer 1

\begin{align} \begin{vmatrix}p^2 & p & 1\\ q^2 & q & 1\\ r^2 & r & 1\end{vmatrix} &=\begin{vmatrix}q^2&q\\r^2&r\end{vmatrix}-\begin{vmatrix}p^2&p\\r^2&r\end{vmatrix}+\begin{vmatrix}p^2&p\\q^2&q\end{vmatrix}\\ &=(q^2r-qr^2)-(p^2r-pr^2)+(p^2q-pq^2)\\ &=q^2r-qr^2-p^2r+pr^2+p^2q-pq^2\\ &=\frac{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}{(r−q)(r−p)(p−q)}(r−q)(r−p)(p−q)\\ &=\frac{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}{q^2r-qr^2-p^2r+pr^2+p^2q-pq^2}(r−q)(r−p)(p−q)\\ &=(1)(r−q)(r−p)(p−q)=(r−q)(r−p)(p−q)\\ \end{align}