Partial Fraction Decomposition of $\frac{x^2}{(x-1)(x+1)^2}$

Question

I am suppose to find the Partial Fraction Decomposition of $$\dfrac{x^2}{(x-1)(x+1)^2}$$

My work so far:

$$\dfrac{x^2}{(x-1)(x+1)^2} = \dfrac{A}{x-1} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$$

Thus, $$ x^2 = A(x+1)^3 + B(x-1)(x+1)^2 + C(x+1)(x-1)$$

With x = 0 we get, $$ 0 = A - B - C $$ I am not sure what to do after this, as plugging in the remaining factors (x = -1, 1) Does not help me find the unknown coefficients.

Any help would be appreciated. Thanks

Edit. With $x = 1$, $$1 = A + 0B + 0C$$ $$A = 1$$

With $x = -1$ $$1 = 0A + 0B + 0C$$ $$1 = 0$$

Answer 1

After your "thus" it must be

$$x^2=A(x+1)^2+B(x-1)(x+1)+C(x-1)$$

and not what you wrote there. Now substitute $\;x=1\;$ in both sides and get

$$1^2=A\cdot2^2\implies A=\frac14$$

and then substitute $\;x=-1\;$ and get ...etc.