Originally Posted by
Reckoner Show us what you have done so far so that we can see what, exactly, you are having trouble with.
Once you have factored the denominator completely, set up your new fractions:
For every linear factor $\displaystyle (ax + b)$ include the term $\displaystyle \frac{A}{ax + b}$. If you have powers of the same factor, $\displaystyle (ax + b)^n$, include all of the terms $\displaystyle \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$.
For each quadratic factor (or power of a quadratic factor), $\displaystyle (ax^2 + bx + c)^n$, include all of the terms $\displaystyle \frac{A_1x + B_1}{ax^2 + bx + c} + \frac{A_2x + B_2}{(ax^2 + bx + c)^2} + \cdots + \frac{A_nx + B_n}{(ax^2 + bx + c)^n}$.
So, in this case we have
$\displaystyle \frac{x^2}{x^3 - 8} = \frac{x^2}{(x - 2)(x^2 + 2x + 4)}$
$\displaystyle =\frac{x^2}{{\color{red}(x - 2)}{\color{blue}(x^2 + 2x + 4)}}$
$\displaystyle ={\color{red}\frac A{x - 2}} + \boxed{{\color{blue}\frac B{x^2 + 2x + 4}}}$
So, we have
$\displaystyle \frac{x^2}{x^3 - 8}=\frac A{x - 2} + \boxed{\frac B{x^2 + 2x + 4}}$
Multiply through by $\displaystyle x^3 - 8$:
$\displaystyle x^2 = A(x^2 + 2x + 4) + B(x - 2)$
Collect terms of the same degree, and equate the coefficients of both sides. Can you finish?