Decompose:

$\displaystyle {x\over {x^4-a^4}}$

answer is:

$\displaystyle {-{{{1\over 2a^2}x}\over {x^2+a^2}}}+{{{1\over 4a^2}\over {x-a}}+{{1\over 4a^2}\over {x+a}}$

I solve for b=$\displaystyle {a\over 4a^3}={1\over {4a^2}}$, however i can't seem to solve for a or c.(if x=-a then -a^2 and -a makes all factors =0)

I try using the quadratic method but i end up with equations that are cubic.