Use the Multinomial Theorem to find the coefficient of x^7*y^8 in
(2*x^3 + 3*x^2*y + 5*x*y^2 + 7*y^3)^5.
the coefficient is $\displaystyle \sum \binom{5}{i,j,k,\ell}2^i3^j5^k7^{\ell},$ where the sum is taken over all integer solutions of the system: $\displaystyle \begin{cases} 3i +2j+k=7 \\ j+2k+3\ell = 8 \end{cases},$ and $\displaystyle 0 \leq i,j,k,\ell \leq 5.$ note
that the condition $\displaystyle i+j+k+\ell=5$ is a result of the above equations. also see that we actually have: $\displaystyle 0 \leq i \leq 2, \ 0 \leq j \leq 3, \ 0 \leq k \leq 4, \ 0 \leq \ell \leq 2.$ one
way to find all solutions is to consider four cases $\displaystyle j=0, 1, 2, 3.$ it will take less than 5 minutes of your time!