# Coefficient problem in algebra

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.!

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 $$\binom{9}{5,4,0}+\binom{9}{6,1,2}(-1)^2=126+252=378.$$

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

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 $$\displaystyle\binom92(1+x^2)^7$$

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

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

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

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