Input:
**User**:
Define
\[p = \sum_{k = 1}^\infty \frac{1}{k^2} \quad \text{and} \quad q = \sum_{k = 1}^\infty \frac{1}{k^3}.\]Find a way to write
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}\]in terms of $p$ and $q.$
Please reason step by step, and put your final answer within \boxed{}.

To evaluate the double sum

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},
$$

we begin by observing that the expression involves a sum over all positive integers $ j $ and $ k $, and the term depends on the sum $ j + k $. This suggests a change of variables to simplify the structure of the sum.

---

### Step 1: Change of Variables

Let us define a new variable $ n = j + k $. For each fixed $ n \geq 2 $, the number of integer pairs $ (j, k) $ such that $ j + k = n $ and $ j, k \geq 1 $ is $ n - 1 $. This is because $ j $ can range from 1 to $ n - 1 $, and $ k = n - j $.

Thus, the double sum can be rewritten as:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = \sum_{n = 2}^\infty \frac{n - 1}{n^3}.
$$

---

### Step 2: Simplify the Sum

We now simplify the expression:

$$
\sum_{n = 2}^\infty \frac{n - 1}{n^3} = \sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right).
$$

This is a telescoping-like sum, and we can split it into two separate sums:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} - \sum_{n = 2}^\infty \frac{1}{n^3}.
$$

Now, recall the definitions:

- $ p = \sum_{n = 1}^\infty \frac{1}{n^2} = \zeta(2) $,
- $ q = \sum_{n = 1}^\infty \frac{1}{n^3} = \zeta(3) $.

So, we can write:

$$
\sum_{n = 2}^\infty \frac{1}{n^2} = p - 1, \quad \sum_{n = 2}^\infty \frac{1}{n^3} = q - 1.
$$

Substituting back, we get:

$$
\sum_{n = 2}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) = (p - 1) - (q - 1) = p - q.
$$

---

### Final Result

Thus, the original double sum simplifies to:

$$
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} = p - q.
$$

---

### Final Answer

$$
\boxed{p - q}
$$

Solution:
We count the number of times $\frac{1}{n^3}$ appears in the sum
\[\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3},\]where $n$ is a fixed positive integer.  (In other words, we are conditioning the sum on $j + k$.)  We get a term of $\frac{1}{n^3}$ each time $j + k = n.$  The pairs $(j,k)$ that work are $(1,n - 1),$ $(2,n - 2),$ $\dots,$ $(n - 1,1),$ for a total of $n - 1$ pairs.  Therefore,
\begin{align*}
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3} &= \sum_{n = 1}^\infty \frac{n - 1}{n^3} \\
&= \sum_{n = 1}^\infty \left( \frac{n}{n^3} - \frac{1}{n^3} \right) \\
&= \sum_{n = 1}^\infty \left( \frac{1}{n^2} - \frac{1}{n^3} \right) \\
&= \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{n^3} \\
&= \boxed{p - q}.
\end{align*}