That is $\ 10^3+1^3=7.11.13$.
I could find no other examples. So I am looking to see if there are any more solutions to $ x^3+y^3=p.q.r$, where $ x, y$ are positive integers and $ p<q<r$ are consecutive odd primes. Now $ x^3 + y^3 =LQ$ where $ L=x+y$, and $ Q = L^2 -3xy$.
After playing with the problem for a while I thought maybe I should try and express the triple of factors in terms of the gaps between them. So $ g_1 = q-p$, and $ g_2=r-q$, and I was thinking what was the biggest possible factor that divided the resulting sum of cubes after choosing the gaps.
Using elementary arguments I found that, $ L=q$, or $ L=r$ , so dealing with the later case put $ L=r$, and $ Q=pq= L^2 -3xy$, eliminate $p$, and $q$, and put $ x = (L+d)/2$, $ y=(L-d)/2$ for $d$ an odd positive integer. Giving $ L^2 - (g_1+2g_2)L+g_1(g_1+g_2)=L^2-3(L^2-d^2)/4$. Completing the square and simplifying gives $ z^2-9d^2 = 4f$, where $ z= 3L-2(g_1+2g_2)$, and $ f = g_1^2+g_1g_2+g_2^2$, the gaps are even so suitable factors can always be found, that is $a$, $b$, such that $4f= ab$, and we get using the difference of squares $z= (a+b)/2$, and so $L=\frac{z+2(g_1+2g_2)}{3}$. Now the largest $z$ is $f+1$, and so the largest,
$$L=\frac{g_1^2+g_1g_2+g_2^2+1+2g_1+4g_2}{3}$$ and chosing $g$ to be the bigger of the two gaps gives a largest $L=\frac{3g^2+6g+1}{3}<(g+1)^2$
so for both cases the largest possible "prime" factor in any triple of solutions was bounded above by roughly the square of the largest gap. Looking at Wikipedia prime gaps then Dr T Nicely's site on first occurrence of prime gaps https://faculty.lynchburg.edu/~nicely/gaps/gaplist.html I noticed that apart from a number of small cases that the primes where gaps first occurred where larger than $(g+1)^2$ suggesting that 1001 is the only case for the range of known first occurrences of prime gaps. I also saw somewhere but cannot remember at the moment something called Shank's conjecture, which is something like the first occurrence of a gap is after a prime that is about $e^\sqrt g$, which is obviously for large enough gap much bigger than $(g+1)^2$. See section 7 Marek Wolf 'Some heuristics on the gaps between consecutive primes' https://arxiv.org/pdf/1102.0481v2.pdf.
So this is roughly where I have got with the problem. I suspect I have missed a simpler solution?
35 is the only sum of two cubes equal to the product of two consecutive odd primes, using the notation above, let $ x^3 +y^3 = pq $, and $ 8<p<q$ be consecutive odd primes then $ L< Q = L^2 -3xy $, Put $ p=L$, $ q=Q$ and $ x=(L+d)/4$, $ y = (L-d)/4$, then $ (p^2+3d^2)/4=q$, so $ (p^2+3)/4 \le q$ but for $p> 8$, $ q>2p$, which contradicts Bertrand's Postulate whereby if we have consecutive primes then $ p< q<2p$ https://en.wikipedia.org/wiki/Bertrand's_postulate
