Equating Coefficients in Partial fractions

Question

I'm having a hard time figuring out A, B, and C for this problem.

$$ \frac{8}{y^{3} - 4y} $$

All I've got so far is

$$ \frac{A}{y} + \frac{B}{(y+2)} + \frac{C}{(y-2)} $$ $$ 8 = A(y+2)(y-2) + B(y)(y-2) + C(y)(y+2) $$

Once you get to this point do you just distribute? After that how do you solve for A, B, C?

Answer 1

Plug in different values for $y$. For instance, plugging in

• $y=0$ gives us $A(2)(-2) = 8 \implies A=-2$
• $y=-2$ gives us $B(-2)(-4) = 8 \implies B=1$
• $y=2$ gives us $C(2)(4) = 8 \implies C=1$

Answer 2

$$A(y^2-4)+B(y^2-2y)+C(y^2+2y)=(A+B+C)y^2+(2C-2B)y-4A=8$$

So,

$$A+B+C=0\\ 2C-2B=0\\ -4A=8.$$