hmmm well, don't be so hmmm off due to it, the wording in the exercise sux0rs bad, is very poorly worded and laid out.
if X and Y are numbers, hmmm say let's give them hmmm ohhh X = 7 and Y = 13.
And X and Y are also complete cubes, well, let's make them so, X = 7³ and Y = 13³.
which of those expressions are complete cubes, or namely, something that we can write as a number with a "3" in the exponent, let's check each one.

what's wrong with the last one? well, if we were to add 91³ + 3³ = 753598. Now, is 753598 a complete cube? well, only if we could write it as a whole number with a "3" above, can we? nope.
we can simply check that by getting the 3rd root of that value,
![\sqrt[3]{753598}\approx 91.00108681228277\impliedby \textit{not a complete cube}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7mwvrha0414quslsgk333kgs0j3ppqzaxd.png)