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?

You have three factors without multiplicity, so try
$$\frac{1}{k(k+1)(k+2)} \equiv \frac{A}{k}+\frac{B}{k+1}+\frac{C}{k+2}$$
Spoiler below

 $$\frac{1}{k(k+1)(k+2)} \equiv \frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}$$

