How to prove $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$? I saw on facebook some image on which these identities that I am going to write below are labeled as "amazing math fact" and on the image there are these identities:

$1^3+5^3+3^3=153$
$16^3+50^3+33^3=165033$
$166^3+500^3+333^3=166500333$
$1666^3+5000^3+3333^3=166650003333$

and then it is written under these identities "and so on and on and on and on!" which suggests that for every $k \in \mathbb N$ we shuld have

$(1 \cdot 10^k + \sum_{i=0}^{k-1} 6 \cdot 10^i)^3 + (5 \cdot 10^k)^3 + (\sum_{i=0}^{k} 3 \cdot 10^i)^3=16...650...03...3$

(on the right hand side of the above stated identity the number of times that number $6$ is shown up is $k-1$, the number of times that number $0$ is shown up is $k-1$ and the number of times that number $3$ is shown up is $k$)
This problem seems attackable with mathematical induction but I would like to see how it could be proved without using mathematical induction in any step(s) of the proof.

This deserves a shorter proof. Call $x=10^{k+1}$. Then we have:
$$\begin{array}{rcl}166\ldots 666&=&\frac{x}{6}-\frac{2}{3}\\500\ldots 000&=&\frac{x}{2}\\333\ldots 333&=&\frac{x}{3}-\frac{1}{3}\end{array}$$
Now: the sum of the cubes on the left-hand side is:
$$\color{red}{166\ldots 666^3+500\ldots 000^3+333\ldots 333^3}=\left(\frac{x}{6}-\frac{2}{3}\right)^3+\left(\frac{x}{2}\right)^3+\left(\frac{x}{3}-\frac{1}{3}\right)^3=\frac{1}{6^3}((x-4)^3+(3x)^3+(2x-2)^3)=\frac{1}{6^3}(36x^3-36x^2+72x-72)=\color{red}{\frac{1}{6}(x^3-x^2+2x-2)}$$
On the other hand, the big number on the right-hand side is:
$$\color{red}{1666\ldots666\cdot x^2+500\ldots000\cdot x+333\ldots333}=\left(\frac{x}{6}-\frac{2}{3}\right)\cdot x^2+\frac{x}{2}\cdot x+\frac{x}{3}-\frac{1}{3}=\frac{1}{6}(x^3-4x^2+3x^2+2x-2)=\color{red}{\frac{1}{6}(x^3-x^2+2x-2)}$$
which is the same value.
(Note how we used multiplication by $x^2$ and $x$ to "shift those big numbers to the left".)
