What is the best way to find determinant of the following matrix?
$$A=\left(\begin{matrix} 1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2 \end{matrix}\right)$$
I thought it looks like a Vandermonde matrix, but not exactly. I can't use $|A+B|=|A|+|B|$ to form a Vandermonde matrix. Please suggest. Thanks.
On
I think the best way it's the following: $$\Delta=\sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=a\sum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$
On
Note that $$ \det\left(\begin{matrix} 1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2 \end{matrix}\right) =\det \left(\begin{matrix} 1&ax&x^2\\1&ay&y^2\\ 1&az&z^2 \end{matrix}\right)=a\cdot\det \left(\begin{matrix} 1&x&x^2\\1&y&y^2\\ 1&z&z^2 \end{matrix}\right) $$ which boils down to Vandermonde determinant $$ a(x-y)(y-z)(z-x). $$
\begin{align} &|A|\\ &=\det\left(\begin{matrix} 1&ax&a^2+x^2\\1&ay&a^2+y^2\\ 1&az&a^2+z^2 \end{matrix}\right) \\ &=\begin{vmatrix} 1&ax&a^2\\1&ay&a^2\\ 1&az&a^2 \end{vmatrix}+\begin{vmatrix} 1&ax&x^2\\1&ay&y^2\\ 1&az&z^2 \end{vmatrix} \tag{multilinearity on 3rd column} \\ &=0+a\begin{vmatrix} 1&x&x^2\\1&y&y^2\\ 1&z&z^2 \end{vmatrix} \\ &= a (x-y)(y-z)(z-x) \end{align}