How to derive this sequence: $1^3+5^3+3^3=153,16^3+50^3+33^3=165033,166^3+500^3+333^3=166500333,\cdots$?

First some notations, let express the repunits $r_n=11\cdots 1=\dfrac{(10^n-1)}9$ in term of $k=10^n$

$\begin{cases} a_n&=16\cdots 6&=10^{n-1}+6\ r_{n-1}&=\frac{k-4}6\\ b_n&=50\cdots 0&=5\times 10^{n-1}&=\frac{k}2\\ c_n&=33\cdots 3&=3\,r_n&=\frac{k-1}3\end{cases}$

The sum of cubes

$f(n)=a_n^3+b_n^3+c_n^3=\left(\frac{k-4}6\right)^3+\left(\frac k2\right)^3+\left(\frac{k-1}3\right)^3=\dfrac{k^3-k^2+2k-2}6=\dfrac{(k-1)(k^2+2)}6$

Can be expressed like this: $$\boxed{f(n)=\dfrac{(10^n-1)(10^{2n}+2)}{6}}$$

But we are interested in the non-factored expression:

$\begin{align}f(n)&=\dfrac{k^3-k^2+2k-2}6\\&=\dfrac{k^3-4k^2+3k^2+2k-2}6\\&=\left(\dfrac{k^2-4k^2}6\right)+\left(\dfrac {3k^2}6\right)+\left(\dfrac{2k-2}6\right)=k^2\,a_n+k\,b_n+c_n=\overline{a_nb_nc_n}\end{align}$

Where

$$\overline{a_nb_nc_n}=\underbrace{16\cdots 6}_{a_n\times 10^{2n}}\;\underbrace{50\cdots 0}_{b_n\times 10^n}\;\underbrace{33\cdots 3}_{c_n}$$

The $n$th such equation is $$\left(\tfrac{5\times 10^{n-1}-2}{3}\right)^3+(5\times 10^{n-1})^3+\left(\tfrac{10^n-1}{3}\right)^3=10^{3n-1}+\tfrac{2}{3}(10^{n-1}-1)10^{2n}+5\times 10^{2n-1}+\tfrac{10^n-1}{3},$$or with $x=10^n$ we can write it as $$\left(\tfrac{x-4}{6}\right)^3+\left(\tfrac{x}{2}\right)^3+\left(\tfrac{x-1}{3}\right)^3=\tfrac{x^3}{10}+\tfrac{2}{3}\left(\tfrac{x}{10}-1\right)x^2+\tfrac{x^2}{2}+\tfrac{x-1}{3},$$which we can verify algebraically (both sides are $\tfrac{(x-1)(x^2+2)}{6}$).