Different ABTwo-Digit Numbers Question

Question

I have seen a question as;

If $\overline{AB}$ is a two-digit number

and

$\overline{AB}-10B-A=A^3 - B^3$

How many different AB two-digit numbers can be written?

Thank you.

Answer 1

$\overline{AB}-10B-A=A^3-B^3\\(10A+B)-10B-A=A^3-B^3\\9(A-B)=(A-B)(A^2+B^2+AB)\\SO\\$ $$\begin{cases}A=B & \rightarrow A=B=1,2,3,4,...9\\A^2+B^2+AB=9 & \begin{cases}A=1 & B^2+1B=9-1\\A=2 & B^2+2B=9-4 \rightarrow \\A=3 & B^2+3B=9-9 \rightarrow B=0\end{cases} \end{cases} \\\overline{AB}=11,22,33,...,99,30$$

Answer 2

$$ \begin{align} &10A+B-10B-A=A^3-B^3\\ \Rightarrow\ &\frac{(A^2+B^2+AB)(A-B)}{A-B}=9\\ \Rightarrow\ &A^2+B^2+AB=9\ or\ A=B \\ \end{align} $$

The solution if $A\ne B$ is $\overline{AB}=30$ i.e. $A=3,B=0$.

For $A=B$, there are $9$ solutions i.e. $11,22,...,99$.

Hence there are $10$ solutions.