Why $3^{3}+4^{3}+5^{3}$ is equal $6^{3}$ and with different numbers length too?

Question

I was playing around with finding roots for an equation like that: $$a_{1}^{n}+a_{2}^{n}+...+a_{n}^{n}=b^{n}, a,b\in \mathbb{Z}$$ and came up with the fact, that $3^{3}+4^{3}+5^{3}=6^{3}$, but not only that, $33^{3}+44^{3}+55^{3}=66^{3}$ is also valid. Actually, as far as I checked, it looks like $$3...3^{3}+4...4^{3}+5...5^{3}=6...6^{3}$$ (where each term has the same length, e.g. $3333^{3}+4444^{3}+5555^{3}=6666^{3}$) is also true.
Now I wonder, why is that?

Answer 1

what you have is: $$x^3+(x+1)^3+(x+2)^3-(x+3)^3=0$$ $$2x^3-12x-18=0$$ $$x^3-6x-9=0$$ and the only real solution to this is $x=3$ hence why it works for $3,4,5,6$. Now notice that: $$3^3=(3\times1)^3=3^3\times1^3$$ $$33^3=(3\times11)^3=3^3\times11^3$$ and the same can be done for $4,5,6$. Hence why any string of $3,4,5,6$ will satisfy this

Answer 2

$\underbrace{33\ldots 3^3}_n+44\ldots 4^3+55\ldots 5^3$ (each digit repeated $n$ times)

$=(3.\sum_{i=0}^{n-1}10^i)^3+(4.\sum_{i=0}^{n-1}10^i)^3+(5.\sum_{i=0}^{n-1}10^i)^3$

$=3^3.\left(\sum_{i=0}^{n-1}10^i\right)^3+4^3.\left(\sum_{i=0}^{n-1}10^i\right)^3+5^3.\left(\sum_{i=0}^{n-1}10^i\right)^3$

$=(3^3+4^3+5^3).\left(\sum_{i=0}^{n-1}10^i\right)^3$

$=6^3.\left(\sum_{i=0}^{n-1}10^i\right)^3$

$=\left(\sum_{i=0}^{n-1}6.10^i\right)^3$

$=66\ldots 6^3$ (n times).