How to prove this curiosity that has to do with cubes of certain numbers?

Consider the three $n$-digit numbers \begin{align} a &= 16\ldots6 = \tfrac16 (10^n - 4), \\ b &= 50\ldots0 = \tfrac12 10^n, \\ c &= 33\ldots3 = \tfrac13 (10^n - 1). \\ \end{align}

Then \begin{align} a^3 + b^3 + c^3 &= \frac{1}{6^3} (10^n - 4)^3 + \frac{1}{2^3} (10^n)^3 + \frac{1}{3^3} (10^n - 1)^3. \end{align} Working out the major terms on the right-hand side,

\begin{align} \frac{1}{6^3} (10^n - 4)^3 &= \frac1{6^3}(10^{3n} - 3\cdot4\cdot 10^{2n} + 3\cdot4^2\cdot 10^n - 4^3) \\ &= \frac{1}{216}10^{3n} - \frac{1}{18}10^{2n} + \frac{2}{9}10^n - \frac{8}{27}, \\[.7ex] \frac{1}{3^3} (10^n - 1)^3 &= \frac1{3^3}(10^{3n} - 3\cdot 10^{2n} + 3\cdot 10^n - 1) \\ &= \frac{1}{27}10^{3n} - \frac{1}{9}10^{2n} + \frac{1}{9}10^n - \frac{1}{27}, \\[.7ex] \frac{1}{6^3} (10^n - 4)^3 + \frac{1}{3^3} (10^n - 1)^3 &= \left(\frac{1}{216} + \frac{1}{27}\right)10^{3n} - \left(\frac{1}{18} + \frac{1}{9}\right)10^{2n} \\ & \qquad + \left(\frac{2}{9} + \frac{1}{9}\right)10^n - \left(\frac{8}{27} + \frac{1}{27}\right) \\ &= \frac{1}{24} 10^{3n} - \frac{1}{6} 10^{2n} + \frac{1}{3}10^n - \frac{1}{3}, \\[.7ex] \frac{1}{2^3} (10^n)^3 &= \frac{1}{8} 10^{3n}. \end{align}

Then, since $\frac{1}{24} + \frac{1}{8} = \frac{1}{6}$, \begin{align} a^3 + b^3 + c^3 &= \frac1{6^3} (10^n - 4)^3 + \frac1{2^3} (10^n)^3 + \frac1{3^3} (10^n - 1)^3 \\ &= \frac16 10^{3n} - \frac16 10^{2n} + \frac13 10^n - \frac13. \end{align} But $$\frac16 10^{3n} = \overbrace{16\ldots6}^{\text{$3n$ digits}}.666\ldots.$$ Subtract $$\frac16 10^{2n} = \overbrace{16\ldots6}^{\text{$2n$ digits}}.666\ldots$$ and the result is $$\frac16 10^{3n} - \frac16 10^{2n} = \overbrace{16\ldots6}^{\text{$n$ digits}}\overbrace{50\ldots0}^{\text{$2n$ digits}}.$$

Continuing, \begin{align} \frac13 10^n &= \overbrace{33\ldots3}^{\text{$n$ digits}}.333\ldots, \\ \frac13 &= \phantom{33\ldots{}}0.333\ldots, \\ \frac13 10^n - \frac13 &= \overbrace{33\ldots3}^{\text{$n$ digits}}, \\ \frac16 10^{3n} - \frac16 10^{2n} + \frac13 10^n - \frac13 &= \overbrace{16\ldots6}^{\text{$n$ digits}}\overbrace{50\ldots000\ldots0}^{\text{$2n$ digits}} \\ & \phantom{=} + \phantom{6\ldots650\ldots0} \overbrace{33\ldots3}^{\text{$n$ digits}} \\[1ex] &= \overbrace{16\ldots6}^{\text{$n$ digits}}\overbrace{50\ldots0}^{\text{$n$ digits}}\overbrace{33\ldots3}^{\text{$n$ digits}}. \end{align}

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".)