Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$. 
Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$. 


I simplified the expression and it turned out to be $(x+3)^7(x+2)^7(x+1)^3$. Now I'm stuck. I only know how to deal with 2 terms but there are 3 over here. I would appreciate a hint. Thanks

Considering $f(x) = (x+3)^7(x+2)^7(x+1)^3$
In expanded form, the sum of roots of this polynomial would be 
$$S = \frac{-b}{a} $$
Here , $b$ would be the coefficient of $x^{16}$, and $a$ the coefficient of $x^{17}$[This is a standard result for any polynomial, that sum of roots is negative of coefficient of second highest power divided by coefficient of highest power (Can be 0)]
Now, $a = 1$ (Easy to check)
Hence, Coefficient of $x^{16} = -S = -((-3)*7+(-2)*7 +(-1)*3) = 38 $
Approach 2
You have $f(x) = (x+3)^7(x+2)^7(x+1)^3$
You can obtain $x^{16}$ by the following cases:
1) $x^7$ from $(x+3)^7$ , $x^7$ from $(x+2)^7$ , $x^2$ from $(x+1)^3$
2) $x^7$ from $(x+3)^7$ , $x^6$ from $(x+2)^7$ , $x^3$ from $(x+1)^3$
3) $x^6$ from $(x+3)^7$ , $x^7$ from $(x+2)^7$ , $x^3$ from $(x+1)^3$
From
1) 7C7*7C7*3C2 ,  
2) 7C6 *(3) *7C7*3C3 , 
3) 7C7*7C6 *(2) *3C3
Adding these , we get Coefficient $= 38$
