Number whose Square and Cube use all Digits Once

Theorem
The only integer whose square and cube use each of the digits from $0$ to $9$ exactly once each is $69$.

Theorem
First the limits are established for the square and cube of an integer to have exactly $10$ digits between them.

So integers smaller than $47$ do not have enough digits available, and integers greater than $99$ have too many.

$47$ and $99$ themselves are eliminated on inspection of the above.

It is noted that integers ending in $0$, $1$, $5$ and $6$ have squares and cubes ending in those same digits.

Such numbers can be eliminated from our search, as they will duplicate the appearance of those digits.

It remains to check the integers between $48$ and $98$.

First the squares which duplicate at least one digit are eliminated:

Of the remainder, the cubes which duplicate at least one digit are eliminated:

Of the remainder, the squares and cubes which duplicate at least one digit between them are eliminated:

Finally:

The result follows.