Evaluating Vandermonde Determinants

Suppose we have four points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$. If we want to find a cubic polynomial $a+bx+cx^2+dx^3$ passing through these points, we have to solve $$\underbrace{\begin{pmatrix}1&x_1&x_1^2&x_1^3\\1&x_2&x_2^2&x_2^3\\1&x_3&x_3^2&x_3^3\\1&x_4&x_4^2&x_4^3\end{pmatrix}}_{=A}\begin{pmatrix}a\\b\\c\\d\end{pmatrix}=\begin{pmatrix}y_1\\y_2\\y_3\\y_4\end{pmatrix}.$$ There is always a unique solution when $A$ is invertible. $A$ is actually a Vandermonde matrix, which is invertible if all the $x_i$'s are distinct. How do we evaluate a Vandermonde determinant? As an example, let's consider $$A=\begin{pmatrix}1&3&9&27\\1&2&4&8\\1&4&16&64\\1&5&25&125\end{pmatrix}.$$ Replace the last row with the variable $x$. Let $$f(x)=\begin{pmatrix}1&3&9&27\\1&2&4&8\\1&4&16&64\\1&x&x^2&x^3\end{pmatrix}.$$ We see that $f(2)=f(3)=f(4)=0$ and $\det(A)=f(5)$. Since $f(x)$ is a cubic polynomial, we must have $f(x)=k(x-2)(x-3)(x-4)$, where $k$ is a constant. Now we perform cofactor expansion along the last row: $$f(x)=1\cdot c_{41}+x\cdot c_{42}+x^2\cdot c_{43}+x^3\cdot c_{44}.$$ Comparing the coefficients of $x^3$ of the two expressions, we have $$k=c_{44}=\begin{vmatrix}1&3&9\\1&2&4\\1&4&16\end{vmatrix}.$$ Do you notice something? This problem is related to recursion! Using the same trick, we can argue that $c_{44}=(4-3)(4-2)(2-3)=-2$. Finally, $f(5)=-2(5-2)(5-3)(5-4)=-12$.

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vandermonde