Binomial Theorem with Three Terms $(x^2 + 2 + \frac{1}{x} )^7$
Find the coefficient of $x^8$
Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.
Does anyone have a method of solving this questions and others similar efficiently?
Thanks.

To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 \choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7\choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 \choose 4}+{7\choose 5}$.
This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have 
7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7
7+4+4, 4+7+4, 4+4+7
6+6+3, 6+3+6, 3+6+6
6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6
That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).
