What's the number?!

why do you Need the solution so urgent?

As Joshua Kimmich mentioned, there would be only $5$ three-digit numbers, namely $X=\{125,216,343,512,729 \}$ . So, $BAD$ can be a memeber of $X$ . Then, we know that a square, mod $4$ is either $0,1$ . therefore, $CBAD \equiv 0,1 (mod \ 4)$ . All members of $X$ pass the test, except $343$ . The new set is $Y=\{125,216,512,729 \}$ . Also note that $A-C$ needs to be an even number, because

$ABCD-CBAD=(A-C)10^3+(C-A)10=10(A-C)(10^2-1)=9\times 10 \times 11(A-C)$

and $9\times 10 \times 11$ has only one $2$ and we know that the difference of squares cannot have only one $2$ .

So, when $A,B,D$ are taken from memebers of $Y$ , for any $BAD$ , you get at most $4$ possible $C$ . So, you are gonna need to take $16$ square roots in the worst case Scenario. Take the square root of $CBAD$ , for a $BAD \in Y$ and a possible $C$ , and if it is a whole number, then test $ABCD$ .

First, I brought up lists of perfect squares and perfect cubes. Next, I noted that all three numbers end in the same number. Therefore, looking at the list of perfect squares and cubes, this last digit had to be 5, 6, or 9. Using the same method, the second digit of both perfect squares had to be 1, 2, 3, 5, or 7. Then I quickly scanned through the list of perfect squares and found 1296. In fact, the only four digit perfect square $ABCD$ where $BAD$ is a perfect cube is 1296 (unless I missed one...), so that was a pretty big giveaway.