There are numbers A and B for which
A/(x - 1) + B/(x + 1) = (x + 8)/(x^2 - 1)
for every number $x\neq\pm1$. Find A-B
Multiply by (x^2 - 1) on both sides gives:
\(A(x + 1) + B(x - 1) = x + 8\\ (A + B)x + (A - B) = x + 8\)
Note that this holds true for any real number x.
Therefore, we can compare the coefficients of x on both sides.
\(\begin{cases}A + B &=& 1\\A - B &=& 8\end{cases}\)
Can you take it from here?