Find the coefficient of u^2 v^9 in the expansion of (2u - 3v + u^2 - v^2)^9

[u(2+u)+v(-3-v) ]^9

\(\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\ \binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\ (-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\\)

To get u^2 n must be 1

\((-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\ =(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\ =9v^8\;[u(u+2)\;][(v+3)^{8}]\\~\\ \)

The constant term for (v+3)^8 = 3^8

So the u^2v^8 term will be in here

\(9v^8\;[u(u+2)\;]3^8\\~\\ 9v^8\;[u^2+2u\;]3^8\\~\\ The \;\;u^2v^8 \text{ term will be}\\~\\ 9*3^8*u^2v^8 = 59049\;u^2v^8\)

LaTex

\binom{9}{n}\;[u(u+2)]^n\;[-v(v+3)]^{9-n}\\~\\

\binom{9}{n}\;[u^n(u+2)^n\;][-v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{n-1}\;\binom{9}{n}\;[u^n(u+2)^n\;][v^{9-n}(v+3)^{9-n}]\\~\\

(-1)^{9-1}\;\binom{9}{1}\;[u^1(u+2)^1\;][v^{9-1}(v+3)^{9-1}]\\~\\

=(-1)^{8}\;*9*\;[u(u+2)\;][v^{8}(v+3)^{8}]\\~\\

9v^8\;[u(u+2)\;]3^8\\~\\

9v^8\;[u^2+2u\;]3^8\\~\\

The \;\;u^2v^8 \text{ term will be}\\~\\

9*3^8*u^2v^8 = 59049\;u^2v^8