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.

Let $$R(x)= \left(x^2+2+{1\over x}\right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= \sum _{k=0}^7 {7\choose k}x^{21-3k}(2x+1)^k$$
Clearly if $21-3k\geq 16$ there is no term with $x^{15}$ so $21-3k\leq 15$ so $k\geq 2$.  
Also if $21-3k\leq 7$ we have no term with $x^{15}$ so $21-3k\geq 8$ so $3k\leq 13$ so $k\leq 4$.
If $k=2$ we have $${7\choose 2}x^{15}(2x+1)^2$$ so the term is $21$
If $k=3$ we have $${7\choose 3}x^{12}(2x+1)^3$$ so the term is $35\cdot 8= 280$
If $k=4$ we have $${7\choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$
so the answer is $301$.
