Let's determine those two-digit numbers which are 3 less than the sum of the cubes of their digits!
From the conditions, the following relationship can be written:

$$
a^{3}+b^{3}=10 a+b+3
$$

where $a$ and $b$ are the digits of a two-digit number. Since this can be at most 99, $a^{3}+b^{3} \leqq 102$, which is only possible if $a, b<4$. From this point, we can start trying out possibilities, that is, by listing all the possibilities, we can find the only correct solution. However, we can continue as follows. Write the initial equation in the following form: $(b-1) b(b+1)=b^{3}-b=a\left(10-a^{2}\right)+3$, for $a=1,2,3$, $(b-1) b(b+1)$ yields 12, 15, and 6, respectively, from which it follows that only the value $1 \cdot 2 \cdot 3=6$ is possible. Thus, $b=2$, and $a=3$ indeed satisfies $32=3^{3}+2^{3}-3$.