Need hint for solving the following problem 
If $f(x^{500}-1)=5x^{1015}+3x^{244}+7x+10$. Find the sum of coefficients of $f(x^5+1)$.

Let $x^{500}-1=y$, then $f(y)=5(y+1)^\frac{1015}{500}+3(y+1)^\frac{244}{500}+7(y+1)^\frac{1}{500}+10$.
But, I don't think that it is a right approach as each term of $f(y)$ gets transformed into an infinite series via binomial expansion, in that  case it is not possible to sum the coefficients.

Another approach(This may be wrong!!,Please check)

Let,$x^{500}-1=x^5+1\implies x^{500}-x^5-1=1$.So,$f(x^5+1)=5(x^{500}-x^5-1)x^{1015}+3(x^{500}-x^5-1)x^{244}+7(x^{500}-x^5-1)x+10(x^{500}-x^5-1)$

So,the sum of coeffecients$=5-5-5+3-3-3+7-7-7+10-10-10=-25$

Please give  some hints.

I think a good hint is to take a look at factor theorem!
