Is it possible to find a perfect cube like 111...11? Can we find a perfect cube like $111...111$(all digits are $1$), apart from the number $1$ itself?
It's easy to prove that there can't be anything like $111...11$ that is a perfect square besides $1$, but how to do this for perfect cube? Are there some new techniques to do this?
 A: (Edited version)
If $$k^3=\frac{10^n-1}9$$
then $$k^3-1=\frac{10^n-10}{9}$$
$$9(k-1)(k^2+k+1)=10(10^{n-1}-1)$$
So if such $k$ exists and both sides are non-zero (i.e. $n\neq1$ and $k\neq1$), it implies that,
First situation:
When either of the factors $k^2+k+1$ or $k-1$ is divisible by $10$,
$$k^2+k+1\equiv 0 \text{ (mod 10) or }k=10m+1$$
Notice $\forall k\ge 0$, $k^2+k+1$ must be odd, so the first condition is not satisfied.
For the second condition, $k=10m+1$, if it is true, then
$$9(10m+1)^3=10^n-1$$
Rearrange and simply and we get
$$100m^3+30m^2+3m=\frac{10^{n-1}-1}{9}.$$
So for every $n$, the premise to get this condition right is to get it correct for every $n-1$. So finally it is equivalent to prove when $n=1$, there exist a positive integer(s) $m$ such that $m(100m^2+30m+3)=0$.
But no real solution except $m=0$.
When $m=0, k=1, n=1$
Second situation:
Only $k-1$ is divisible by 5. But then $k^2+k+1$ is divisble by 2, which is impossible.
Third situation:
Only $k-1$ is divisible by 2. We write $k=2t+1$. So we investigate whether $k^2+k+1$ is divisible by 5.
Observe
$$(2t+1)^2+(2t+1)+1=4t^2+6t+3$$
and
$$4t^2+6t+3\equiv -t^2+t-2\pmod{5}$$
This is equivalent to state that
$$t^2-t+2\equiv 0 \pmod{5}$$
But by inspection, this is not valid. (By testing the case 5q, 5q+1, ... , 5q+4)
So this situation is impossible.
