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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.

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    $\begingroup$ Did you mean $\vert A + B \vert = \vert A \vert + \vert B \vert$? $\endgroup$ – Bladewood Feb 15 at 22:20
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\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}

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    $\begingroup$ i am confused which answer to be accepted ....all are nice.... $\endgroup$ – neelkanth Feb 15 at 16:41
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    $\begingroup$ @neelkanth I suggest you throw a dice. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 15 at 16:42
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    $\begingroup$ I rolled I dice three time for better .... got three different choices every time . $\endgroup$ – neelkanth Feb 15 at 16:47
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    $\begingroup$ But that requires more resources. :P So, you save time but you have to have multiple dice then. :P $\endgroup$ – stressed out Feb 15 at 22:58
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    $\begingroup$ @stressedout You're right. That's why I say if possible. Perhaps OP was busy with a simulation of the Law of Large Number that he couldn't decide which answer to accept. :xd $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 15 at 23:01
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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 $$ a(x-y)(y-z)(z-x). $$

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I think the best way it's the following: $$\Delta=\sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=a\sum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$

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