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Find the coefficient of $ x^{8} $ in the expansion of $ (1+x^2-x^3)^{9} $ I know the problem is simple if we use multinomial theorem and I got an answer $ 378 $ using it. Can someone check it and also provide a shorter method if possible.!

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That is correct. Multinomial theorem is the shortest proof here, the point being only 4 $x^2$s or one $x^2$ and two $x^3$ (the rest being $1$s) can give a product $x^8$. So you can almost just write down the answer $$ \binom{9}{5,4,0}+\binom{9}{6,1,2}(-1)^2=126+252=378. $$

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  • $\begingroup$ Thanks! Is there any other way though ? $\endgroup$ – Aditya Garg May 25 at 7:59
  • $\begingroup$ At precalc level? Expanding this is laborious but it does work. Same goes with repeated use of binomial theorem. $\endgroup$ – user10354138 May 25 at 8:06
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Using Binomial Theorem

the coefficient of $x^8$ in $\displaystyle((1+x^2)-x^3)^9$

$=$ the coefficient of $x^8$ in $\displaystyle\binom90(1+x^2)^9+$ the coefficient of $x^2$ in $\displaystyle\binom92(1+x^2)^7$

$\displaystyle=\binom94+\binom92\binom71$

$\displaystyle=126+36\cdot7=?$

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You may reduce it to using the binomial theorem only as follows:

\begin{eqnarray*}[x^8](1+x^2-x^3)^{9} & = & [x^8](1+x^2(x-1))^{9} \\ & = & [x^8]\sum_{k=0}^9 \binom{9}{k}x^{2k}(x-1)^k\\ & = & [x^8]\sum_{k=\color{blue}{3}}^{\color{blue}{4}}\binom{9}{k}x^{2k}(x-1)^k\\ & = & \binom{9}{3}[x^2](x-1)^3 + \binom{9}{4}[x^0](x-1)^4\\ & = & 3\binom{9}{3} + \binom{9}{4}\\ & = & 378\\ \end{eqnarray*}

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