Any neat way to calculate this Vandermonde-like determinant?

Solution 1:

Hint:

$$\begin{bmatrix}1&1&1\\x_1&x_2&x_3\\x_1^2&x_2^2&x_3^2\end{bmatrix}\begin{bmatrix}1&x_1&x_1^2\\1&x_2&x_2^2\\1&x_3&x_3^2\end{bmatrix}=\begin{bmatrix}1+1+1&x_1+x_2+x_3&x_1^2+x_2^2+x_3^2\\x_1+x_2+x_3&x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3\\x_1^2+x_2^2+x_3^2&x_1^3+x_2^3+x_3^3&x_1^4+x_2^4+x_3^4\end{bmatrix}$$

Solution 2:

Another way to the formula: I think that you have almost finished the job.

I consider your formula with the $v_k$. Replace them by $w_1,..,w_n$ any vectors in $E=\mathbb{R}^n$. Your formula give then an application $f(w_1,..w_n)$ from $E^n$ to $\mathbb{R}$, and you have see that this is $a{\rm det}(w_1,...,w_n)$ for a constant $a$, depending on the $x_k$, but not on the $w_k$. Now to find $a$, you can take for the $w_k$, the vectors of the canonical basis of $E$, and it is easy to finish.