# 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?

• Note that cube roots are easier than square roots using this trick, because the final digits for perfect squares are not unique: $4^2=16$ and $6^2=36$ both end in $6$, for example. This is related to the fact that the exponent $2$ has a factor in common with $10$ (the base for the decimal system). For cubes, $3$ and $10$ are mutually prime. You can conclude that there would be a similar trick for seventh powers, but then you'd need to memorize $1^7=1, 2^7=128, 3^7=2187, \ldots$. – Théophile Feb 11 '15 at 1:07

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 $57 = \sqrt[3]{185193}.$

Write the cube root as $10a + b$, $a$ and $b$ being the digits. $$(10a)^3 = 1000a^3$$ Considering the coefficient of $1000$ for our cube ($185$ in your example), and then simply looking that up in a table will give you the coefficient of $10$ in the decimal representation of the root, i.e. its first digit ($5$ in your example).

Now towards the second (last) digit:

$$(10a + b)^3 \equiv c \pmod{10}$$ $$10(100a^3 ~+~ ...) + b^3 \equiv c \pmod{10}$$ $$b^3 \equiv c \pmod{10}$$

All we need for the last digit of our cube-root is the last digit of the cube itself. Fortunately the last digit of the cube of all numbers from $0$ to $9$ is distinct, thus we can easily work out the last digit of the root by reading the following table right to left:

$$0^3 \rightarrow 0$$ $$1^3 \rightarrow 1$$ $$2^3 \rightarrow 8$$ $$3^3 \rightarrow 7$$ $$4^3 \rightarrow 4$$ $$5^3 \rightarrow 5$$ $$6^3 \rightarrow 6$$ $$7^3 \rightarrow 3$$ $$8^3 \rightarrow 2$$ $$9^3 \rightarrow 9$$

The fact that for some cubes the last digit doesn't change is not vital for this method, just as the fact that for all others the last digit is $10$ minus the original last digit. You could still perform the lookup for the second digit if that was not the case.

So you're essentially doing two lookups that, due to the way the decimal system works, suffice to uniquely determine the cube root.