Binomial coefficient after expansion.

Question

What is the coefficient of $x^7$ in the expansion of $(1-x-x^2+x^3)^6$ ?

Answer 1

I think the fastest way to compute this, after observing that $1$ and $-1$ are both obviously roots of $1-x-x^2+x^3$, and that $(1-x)(1+x)=1-x^2$, is to write $1-x-x^2+x^3=(1-x^2)(1-x)$ so that $$ (1-x-x^2+x^3)^6 = (1-x^2)^6 (1-x)^6 = \biggl(\sum_{i\geq0}(-1)^i\binom6ix^{2i}\biggr) \biggl(\sum_{j\geq0}(-1)^j\binom6jx^j\biggr). $$ The coefficient of $x^7$ can only involve odd indices $j$; more precisely it comes from the terms with $(i,j)\in\{(1,5),(2,3),(3,1)$}, and is equal to $$ +\binom61\binom65-\binom62\binom63+\binom63\binom61 = 6\times6-15\times20+20\times6 = -144. $$

Answer 2

Just observe that $$ (1-x-x^2+x^3)^6 = (x+1)^6 (x-1)^{12}. $$ The rest is easy.

Answer 3

Use the multinomial theorem, here: $$ (a+b+c+d)^n = \sum_{k_1 + k_2 + k_3 + k_4 = n} \binom{n}{k_1, k_2, k_3, k_4} a^{k_1} b^{k_2} c^{k_3} d^{k_4} $$ The multinomial coefficient is here: $$ \binom{n}{k_1, k_2, k_3, k_4} = \frac{n!}{k_1! \, k_2! \, k_3! \, k_4!} $$ Now you need to figure out which combinations of terms give the required power of $x$.