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Calculate below determinant by using product of determinants:

$\left[\begin{array}{ccc}(a_o+b_0)^n & ... & (a_0+b_n)^n\\ \vdots & \ddots& \vdots\\(a_n+b_0)^n & ... & (a_n+b_n)^n\end{array}\right]$

Can someone help me?

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$\left[\begin{array}{ccc}(a_0+b_0)^n & ... & (a_0+b_n)^n\\ \vdots & \ddots& \vdots\\(a_n+b_0)^n & ... & (a_n+b_n)^n\end{array}\right]=\left[\begin{array}{ccc}1 & ... & a_0^n\\ \vdots & \ddots& \vdots\\1 & ... & a_n^n\end{array}\right]\left[\begin{array}{ccc}\binom{n}{0}b_0^n & ... & \binom{n}{0}b_n^n\\ \vdots & \ddots& \vdots\\\binom{n}{n}1 & ... & \binom{n}{n}1\end{array}\right]=AB$

$A$ is a Vandermonde matrix and you can find $|A|$ at wikipedia similarly $|B|$ is $\Pi_{i=0,...n}\binom{n}{i}$ times of the determinant of Vandermonde matrix $B^{'}=\left[\begin{array}{ccc}b_0^n & ... & b_n^n\\ \vdots & \ddots& \vdots\\1 & ... & 1\end{array}\right]$

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  • $\begingroup$ You're genius. Thanks a lot. $\endgroup$ – amoneth Jan 18 '14 at 21:52

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