Given the problem : find the coefficient of $x^{100}$ in $\frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}$, why this solution works? We will use partial fractions so :
$\\ \frac{1}{\left(1-x^5\right)\left(1-x^{10}\right)}=\frac{1}{\left(1-x^5\right)(1-x^5)\left(1+x^5\right)}=\:\frac{A}{1-x^5}+\frac{B}{\left(1-x^5\right)^2}+\frac{C}{1+x^5}$
We get
$a\left(1-x^5\right)\left(1+x^5\right)+b\left(1+x^5\right)+c\left(1-x^5\right)^2=1$
Now in the solution my friend did this :
$(*)\\x=1\implies b=\frac{1}{2}\:;\\x=-1\implies c=\frac{1}{4}\:;\\x=0\implies a+b+c=1\implies a=\frac{1}{4}\:;$ 
Why this works? why can I put different values for x?
The rest of the solution I understand :
We need to set i=j=k=20 for the coefficients in 
$\frac{1}{4}\sum \:_{i=0}^{\infty \:}x^{5i}\:+\:\frac{1}{2}\sum \:\:_{j=0}^{\infty \:\:}\left(j+1\right)x^{5j}\:+\frac{1}{4}\sum \:\:_{k=0}^{\infty \:\:}\left(-1\right)^kx^{5k}$
To get
$\frac{1}{4}\cdot 1\:+\:\frac{1}{2}\cdot 21\:+\:\frac{1}{4}\cdot \left(-1\right)^{20}\:=\:11$
I solved it in 2 different ways. But in this way above, why can I do what I did in $(*) ?$

example - 
$(1-x^5)^2$ will be zero when $x=1$
if you multiply by $(1-x^5)^2$ on both sides of the equation (considering your second and third terms),  you then have
$\frac{1}{1+x^5} = A(1-x^5) + B + \frac{C(1-x^5)^2}{1 + x^5}$
if you then go back to your case where x=1, the first and third terms on the right vanish (since they become zero), conveniently leaving
$\frac{1}{2} = B$
If the value of B 'works' when $x=1$, then so long as our partial fraction technique is sound, it has to work for all general values too.
