Why does this cube root trick work? So, I found that you can work out cube roots between $1$ and $100$ using this method:
For example the number: $185193$
If the last digit $x$ is $2$, $8$, $3$, or $7$, the last digit in the answer should be $10 - x$; if otherwise, then $x$. In this case:
$$10-3 = 7$$
We now got $\_7$. Remove the last three digits so that we get only $185$. Then, check in this list:
$$
1^3 = 1 \\
2^3 = 8 \\
3^3 = 27 \\
4^3 = 64  \\
5^3 = 125 \\
6^3 = 216 \\
7^3 = 343 \\
8^3 = 512 \\
9^3 = 729 
$$
Check which two cubes that number is between, the lowest one should be the first digit in the answer:
$$5^3 < 185 < 6^3$$
Therefore, we get:
$$57$$
I found this method in this video: https://www.youtube.com/watch?v=Al7PllobwR4
But how does this method really work, and why?

If $10n \leq x < 10(n+1)$, then $1000n^3 \leq x^3 < 1000(n+1)^3$,
and therefore $$n^3 \leq \frac{x^3}{1000} < (n+1)^3.$$
Moreover, since $n^3$ and $(n+1)^3$ are integers you can ignore the fractional
part of $\frac{x^3}{1000},$ which is essentially what you do when you
drop the last three digits of $x$.
For a two-digit number there are only a few choices for $n$;
so by examining your table of values of $n^3$ you can figure
out which value of $n$ makes the equation above true.
So the second half of your "trick" will correctly compute the tens digit
of the cube root of any positive integer less than one million.
The first half of the trick depends on the assumption that the number you
start with is a perfect cube.  It relies on the fact that in your table of
cubes of a single digit, no two results on the right-hand side have the
same last digit. So if $b$ is a digit and you know the last digit of $b^3$,
you can tell what $b$ is. (The trick gives a mysterious rule for computing
$b$, but all it really has to do is to give the correct answer for
nine different input values. Ten, if you include zero.)
The trick also relies on the fact that
$(10a + b)^3 = 10(100a^3 + 30 a^2 b + 3ab^2) + b^3.$
That is, the last digit of $(10a+b)^3$ is just the last digit of $b^3$.
The trick fails for numbers that are not perfect cubes to begin with.
The first digit will be correct but not the rest.
For example, the trick says the cube root of $185194$ is $54$, but
you know it has to be greater than $57 = \sqrt[3]{185193}.$
