Of Cubes and Squares

First, since the total number of digits in $n^2$ and $n^3$ are $10$ , one can quickly discern that $n$ is a two digit number, $n^2$ is a $4$ digit number and $n^3$ is a $6$ digit number. Also, $n^2$ and $n^3$ cannot both start with $9$ . This would limit $n$ to $96 \ge n \ge 47$ .

Since every digit appears exactly once, note that if $S(n)$ denoted the sum of digits of $n$ then $n^2+n^3 \equiv S(n^2)+S(n^3) \equiv 0+1+2+3+\dots+9 \equiv 0 \pmod 9$ Thus we discern that $n^3+n^2 \equiv 0 \pmod 9$ , and thus $n \equiv 0 \pmod 3$ or $n \equiv 8 \pmod 9$

Also, $n$ cannot contain $0, 1, 5, 6$ as their last digit.

This would mean the only possible values for $n$ are $53, 62, 89$ and $48, 54, 57, 63, 69, 72, 78, 84, 87, 93$ .

Since $53^3$ contains two $7$ s, $62^2$ contains two $4$ s, $89^3$ contains two $9$ s, $48^3$ s contains two $1$ s, $54^3$ contains two $4$ s, $57^3$ contains two $1$ s, $63^3$ contains two $0$ s, $72^3$ contains two $3$ s, $78^3$ contains two $4$ s, $87^3$ contains two $5$ s, these are all impossible.

So the cases remaining are $69, 84$ and $93$ . Since $93^2$ and $93^3$ both start with $8$ ,and because both $84^2$ and $84^3$ contain a $5$ , these two are not the answer.

$69^2=4761, 69^3=328509$ so we conclude that $69$ is the one and only answer.

I admit this is not the nicest answer. It was one of the problems in my old math workbook, and because there was not solution, I had to add my own.

I could only reduce the infinite set of integers as possible answers to $47$ to $94$ excluding numbers containig zero ,1,5,6 as their last digit.