Okay, so I need to find a way to express the double sum \(\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}\) in terms of \(p = \sum_{k = 1}^\infty \frac{1}{k^2}\) and \(q = \sum_{k = 1}^\infty \frac{1}{k^3}\). Hmm, let's start by understanding the problem.

First, the given series \(p\) and \(q\) are the sums of reciprocal squares and cubes respectively. The target is a double sum where the denominator is the cube of the sum of two indices. So, maybe there's a way to manipulate the double sum to relate it to \(p\) and \(q\).

Let me write down the target sum again:

\[
\sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{1}{(j + k)^3}
\]

I wonder if there's a way to change variables here. Let me set \(n = j + k\). Then, for each fixed \(n\), the possible pairs \((j, k)\) such that \(j + k = n\) are \(j = 1, 2, ..., n - 1\) and \(k = n - 1, n - 2, ..., 1\). Wait, but since both \(j\) and \(k\) start at 1, the minimum value of \(n\) is 2 (when \(j = 1, k = 1\)), and it goes to infinity. So, maybe I can rewrite the double sum as a single sum over \(n\) from 2 to infinity, and for each \(n\), the number of terms is \(n - 1\), since \(j\) can be from 1 to \(n - 1\), and \(k = n - j\). 

So, rewriting the double sum:

\[
\sum_{n = 2}^\infty \sum_{j = 1}^{n - 1} \frac{1}{n^3}
\]

But wait, the inner sum is over \(j\) from 1 to \(n - 1\), but the term \(\frac{1}{n^3}\) doesn't depend on \(j\). So, the inner sum is just \((n - 1) \cdot \frac{1}{n^3}\). Therefore, the entire sum becomes:

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

Ah, that's a nice simplification! So, the original double sum is equal to the sum from \(n = 2\) to infinity of \(\frac{1}{n^2} - \frac{1}{n^3}\). Then, this can be written as:

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

Because we're starting the sum from \(n = 2\), which is equivalent to subtracting the \(n = 1\) term from the total sum. So, substituting back in terms of \(p\) and \(q\):

\[
\left( p - 1 \right) - \left( q - 1 \right) = p - q
\]

Wait, so does that mean the original double sum is equal to \(p - q\)? Let me verify this step by step to make sure I didn't make a mistake.

First, let's check the substitution:

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

Change variables: \(n = j + k\), so for each \(n \geq 2\), the number of pairs \((j, k)\) such that \(j + k = n\) is \(n - 1\). Therefore, the sum becomes:

\[
\sum_{n=2}^\infty \sum_{j=1}^{n - 1} \frac{1}{n^3} = \sum_{n=2}^\infty \frac{n - 1}{n^3}
\]

Yes, that seems right. Then, splitting the fraction:

\[
\frac{n - 1}{n^3} = \frac{n}{n^3} - \frac{1}{n^3} = \frac{1}{n^2} - \frac{1}{n^3}
\]

So, the sum becomes:

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

Which is equal to:

\[
\left( p - \frac{1}{1^2} \right) - \left( q - \frac{1}{1^3} \right) = (p - 1) - (q - 1) = p - q
\]

Yes, that seems correct. Therefore, the original double sum is equal to \(p - q\). Let me check with a specific example to ensure this isn't a coincidence.

Suppose we take a finite case, say up to \(j, k \leq 2\). Then the double sum would be:

For \(j=1, k=1\): \(1/8\)
\(j=1, k=2\): \(1/27\)
\(j=2, k=1\): \(1/27\)
\(j=2, k=2\): \(1/64\)

Total sum: \(1/8 + 1/27 + 1/27 + 1/64\)

Calculating each term:

\(1/8 = 0.125\)

\(1/27 \approx 0.037037\)

\(1/64 \approx 0.015625\)

So total sum: \(0.125 + 0.037037 + 0.037037 + 0.015625 \approx 0.2147\)

Now, if \(p = \pi^2/6 - 1 \approx 1.6449 - 1 = 0.6449\), and \(q = \zeta(3) \approx 1.20206\). Then \(p - q \approx 0.6449 - 1.20206 = -0.55716\), which is negative, but our finite sum is positive. Wait, that can't be right. Wait, hold on, maybe I confused the definitions.

Wait, hold on. The given \(p = \sum_{k=1}^\infty \frac{1}{k^2} = \zeta(2) = \pi^2/6 \approx 1.6449\), and \(q = \zeta(3) \approx 1.20206\). Then \(p - q \approx 0.4428\). However, our finite sum is approximately 0.2147. So, in this case, the finite sum is less than \(p - q\). But according to our formula, the infinite sum should be \(p - q\). So, if we take more terms, the finite sum should approach \(p - q\). Let's check with more terms.

But since the original double sum is over all \(j, k \geq 1\), the finite case up to \(j, k \leq 2\) is just a partial sum. So, adding more terms would get closer to \(p - q\). Let's try a larger finite case, say up to \(j, k \leq 3\):

Compute all terms where \(j, k\) from 1 to 3:

\(j=1, k=1\): \(1/8\)

\(j=1, k=2\): \(1/27\)

\(j=1, k=3\): \(1/64\)

\(j=2, k=1\): \(1/27\)

\(j=2, k=2\): \(1/64\)

\(j=2, k=3\): \(1/125\)

\(j=3, k=1\): \(1/64\)

\(j=3, k=2\): \(1/125\)

\(j=3, k=3\): \(1/216\)

Total sum: \(1/8 + 2/27 + 2/64 + 1/125 + 1/216\)

Calculating:

\(1/8 = 0.125\)

\(2/27 \approx 0.074074\)

\(2/64 = 0.03125\)

\(1/125 = 0.008\)

\(1/216 \approx 0.00463\)

Total: \(0.125 + 0.074074 + 0.03125 + 0.008 + 0.00463 \approx 0.24295\)

Which is closer to \(p - q \approx 0.4428\), but still not there yet. So, as we take more terms, the sum approaches \(p - q\). Therefore, the formula seems to hold in these finite cases, which gives me more confidence.

Therefore, the original double sum is indeed equal to \(p - q\).

But let me check again if there's a miscalculation in the steps. When we changed variables to \(n = j + k\), we considered all pairs \((j, k)\) such that \(j + k = n\), which gives \(j\) from 1 to \(n - 1\), so \(n - 1\) terms. Then, each term contributes \(1/n^3\), so the inner sum is \((n - 1)/n^3\). Then, splitting that into \(1/n^2 - 1/n^3\), and summing from \(n = 2\) to infinity.

Then, the sum of \(1/n^2\) from \(n = 2\) is \(p - 1\), and the sum of \(1/n^3\) from \(n = 2\) is \(q - 1\). Therefore, the difference is \(p - 1 - (q - 1) = p - q\). So, yes, that seems correct.

Therefore, the answer is \(p - q\), which in boxed form is \(\boxed{p - q}\).
Therefore, the given double sum can be written in terms of \(p\) and \(q\) as \(\boxed{p - q}\).
</think>