<s>[INST] partial fraction decomposition: product in denominator A fraction $\dfrac{a}{bc}$ can be split into $\dfrac{x}{b} + \dfrac{y}{c}$ by solving for $x$ and $y$ from $a = by + cx$.
Now, then a term $\dfrac{a}{cd}$ should be able to be split like this
$$\frac{x}{bc} + \frac{y}{d}$$
But apparently that is not the case, as I have redone the decomposition for the expression $\dfrac{1}{k(k + 1)(k + 2)}$ several times now:
\begin{align*}
  \frac{1}{k(k + 1)(k + 2)} &=
  \frac{\overbrace{1}^a}{\underbrace{(k^2 + k)}_b\underbrace{(k + 2)}_c}\\
  &= \frac{x}{(k^2 + k)} + \frac{y}{(k + 2)}&\\
  &= \frac{x(k + 2)}{(k + 2)(k^2 + k)} + \frac{y(k^2 + k)}{(k + 2)(k^2 + k)}
\end{align*}
Solving for $x$ and $y$:
\begin{align*}
            & 1 = x(k + 2) + y(k^2 + k)\\
            & k := -2 \implies y = \frac{1}{2}\\
            & 1 = x(k + 2) + \frac{1}{2}(k^2 + k)\\
            & k := -1 \implies x = 1
          \end{align*}
Thus:
$$\frac{1}{k(k + 1)(k + 2)} = \frac{1}{k(k + 1)} + \frac{\frac{1}{2}}{(k + 2)}\\$$
But that is not correct, as this example shows:
$$\frac{1}{3(3 + 1)}+ \frac{\frac{1}{2}}{3 + 2} = \frac{11}{60}$$
while
$$\frac{1}{3(3 + 1)(3 + 2)} = \frac{1}{60}$$
I hope I have not made some silly obvious arithmetic or algebraic mistake, but I have redone this several times, leading me to assume I am making a wrong assumption about how this works. So, is my assumption that there are $x$ and $y$ such that $\dfrac{a}{bcd} = \dfrac{x}{bc} + \dfrac{y}{d}$ wrong?
 [/INST] Your procedure is ok till $1 = x(k + 2) + y(k^2 + k)$.
Now in solving this you make the wrong assumption that $x$ is independent from $k$ (is constant for whichever $k$). 
The theory of partial fractions decomposition says that if the denominator of one the decomposition fractions is (or is kept) of degree 2, then the numerator will, in general, be of degree 1.
The answer above shows you the details.
</s>