Find coefficient for $x^{10}$ in $x^3(x^2-3x^3-1)^6$ I'm trying to solve the following problem:

Find the coefficient for  $x^{10}$ in $x^3(x^2-3x^3-1)^6$.

Can I use the multinomial theorem to solve it? I'm unsure how to start..
Thanks!
 A: After simplifying by $x^3$, you need to solve the Diophantine system 
$$\begin{cases}a+b+c&=6,\\2a+3b+0c&=7.\end{cases}$$
The only solution is $a=2,b=1,c=3.$
Then the requested coefficient is, by the multinomial formula,
$$\binom{6!}{2!1!3!}1^2(-3)^1(-1)^3=180.$$
A: Definition: 
$[x^n]f(x)~$ means the coefficient of $~x^n~$ of the Taylor expansion of $~f(x)~$ at $~x=0~$ . 
Calculation:
$\displaystyle [x^{10}]x^3(x^2-3x^3-1)^6 = [x^{7}](x^2-3x^3-1)^6 = [x^{7}]\sum\limits_{k=0}^6 {\binom 6 k}x^{2k}(3x-1)^k = $
$\displaystyle =\sum\limits_{k=0}^6 {\binom 6 k}[x^{7-2k}](3x-1)^k =\sum\limits_{~~~~~k=0\\0\leq 7-2k\leq k}^6 {\binom 6 k}[x^{7-2k}](3x-1)^k$
$\displaystyle =\sum\limits_{k=3}^3 {\binom 6 k}[x^{7-2k}](3x-1)^k = {\binom 6 3}[x^{1}](3x-1)^3 =20\cdot 3{\binom 3 1}=180$
A: Well it's simple
we are multiplying every term by $x^3$ so coefficient of $x^{10}$ should be coeffiecent of $x^7$ in the expansion of $(x^2-3x^3-1)^6$
Now, divide three terms into two groups $((x^2-3x^3)+(-1))^6$ 
To get $x^7$, the only combination possible is when $x^2$ is raised to the power two and $-3x^3$ is rased to the power one and multiplied.So we get $6C3*(x^2-3x^3)^{(2+1)}*(-1)^{(6-(2+1))}$. I took 3 (2+1) because thats when i get the terms $x^7$.. for $x^7$ we get $6C3*3C2*(x^2)^2*(-3x^3)*(-1)^3$.Now multiply all the coefficent we get $20*3*-3*-1=180$
And you got your answer
