This would have been a nice 1952's problem

Suppose that $3^{3x} - 3^{2x+1} + 3^{x+1} - 1952 = n^3$ for some integer $n.$ Then we can write

$3^{3x} - 3 \cdot 3^{2x} + 3 \cdot 3^x - 1952 = n^3.$

Now, the critical step is to notice this factorization:

$(3^x-1)^3 - 1951 = n^3.$

(Notice the $1, -3, 3$ coefficients on the left-hand side, which was my primary motivation.) Then, rearranging gives $(3^x-1)^3 - n^3 = 1951.$ Let $a = 3^x - 1,$ so we have $a^3 - n^3 = 1951$ or

$(a-n)(a^2+an+n^2) = 1951.$

Normally, we would now proceed to try a bunch of factor pairs of $1951$ ... but we hope, hope, hope that Daniel was nice to us when writing this problem and won't make us try too many. Indeed, $1951$ is prime, the easiest of all! :P

Also, note the following lemma:

Lemma. For all real numbers $a, n$ (whether negative or positive), the following inequality holds: $a^2+an+n^2 \ge 0.$

Proof. If $a, n$ have the same sign (or one/both of them is zero), then $an \ge 0,$ so

$a^2+an+n^2 \ge a^2+n^2 \ge 0+0 = 0.$

Otherwise, if $a, n$ have the opposite sign, then $an \le 0$ ; but now we can write

$a^2+an+n^2 = (a+n)^2 - an \ge (a+n)^2 \ge 0.$

Thus, in all cases, $a^2+an+n^2 \ge 0.$

So, by the Lemma, there are only two cases to try:

(1) Suppose that $a-n=1$ and $a^2+an+n^2 = 1951.$ We substitute $a = n+1$ into the second equation to get $(n+1)^2 + n(n+1) + n^2 = 1951,$ which simplifies to

$3(n^2+n-650) = 0.$

This has roots $n = 25$ or $n = -26.$ But if $n = -26,$ then $a = -25$ and the equation $3^x - 1 = -25$ has no real solutions. Thus, $n = 25,$ so $a = 26,$ so $3^x-1=26 \Longrightarrow x = 3.$ The solution $x = 3$ corresponds to the value $25,$ since we defined $n$ to be the value of the expression.

(2) Suppose that $a-n=1951$ and $a^2+an+n^2 = 1.$ We substitute $a = n+1$ into the second equation to get $(n+1951)^2 + n(n+1951) + n^2 = 1,$ which simplifies to

$3 n^2+5853 n+3806400 = 0.$

This is a mess, and has no real roots for $n,$ so there are no solutions in this case.

So, the only value of $x$ is $x=3,$ which corresponds to the value $\boxed{25}.$ $\square$

For x>3 we have $(3^x-1)^3-(3^x-2)^3>1951$ .That leaves x=3,2,1.Done :D

I think the reason this problem would have been tough is if you were using a crappy calculator. Because I noticed for almost all $x$ 's, the formula produced a number that was like, insanely close to a whole number, like at $x=12$ , the result was $531439.99999999769735209398753712937322597604322742013$ . But I just kept plugging in values. The only real "trick" I used was realizing that $x$ had to be greater than $2$ because otherwise the part under the radical would be negative. Plugging in $x=3$ produces $\boxed{25}$ which I guess is the only truly integral solution. I'll be waiting for a good solution though, @Daniel Liu . Nice problem! :D