Is there a shortcut for finding cube of a particular number like $68^3$ ? If anyone knows how to solve for two- and three digit numbers, can you please share the answer?

  • $\begingroup$ there's a shortcut? $\endgroup$ – Gregory Grant Jul 26 '16 at 10:59
  • $\begingroup$ I found some links by googling: youtube.com/watch?v=FktRm6Ts8w0 and burningmath.blogspot.com/2013/10/… $\endgroup$ – Gregory Grant Jul 26 '16 at 11:03
  • 2
    $\begingroup$ Hint : Use the expansion of $(70-2)^3$ $\endgroup$ – Peter Jul 26 '16 at 11:15
  • $\begingroup$ Another hint: Notice that the linked method above, as well as the solutions posted by callculus and me, all employ the 3rd row of Pascal's Triangle: 1 3 3 1 $\endgroup$ – Grey Matters Jul 28 '16 at 17:30
  • 1
    $\begingroup$ See the book " The Trachtenberg Speed System Of Basic Mathematics" by Jakow Tractenberg. Or the Wikipedia article about it. $\endgroup$ – DanielWainfleet Jul 28 '16 at 22:24

You can apply the binomial theorem for a more comfortable calculation:

$$(a+b)^n=\sum_{k=0}^n~{n\choose k} \cdot a^{n-k} \cdot b^k$$

First note that $68=70-2$

Therefore in your case it is

$$(70-2)^3=\sum_{k=0}^3~{3\choose k} \cdot 70^{3-k} \cdot (-2)^k$$

$=1\cdot 70^3\cdot 1+3\cdot 70^2\cdot (-2)+3\cdot 70\cdot (-2)^2+1\cdot 1 \cdot (-2)^3$

$=343,000-6\cdot 4900+12\cdot 70-8$


In this case it can be calculated without using a calculator.


Trying to see how I think the following: if the number has small prime factors, you can first factoring them and then apply the binomial theorem appropriately with respect to a multiple of 10, or calculate first the square of large prime factors. For example for your $68$ one can do

$$68=2^2\cdot 17\rightarrow68^3=2^6\cdot17^3=2^6\cdot17(20-3)^2$$

Perhaps for three digit numbers may be convenient to write instead of $abc$ the expression $$abc=10ab+c$$ after factorization (when it is easy of course, such as the $2^2$ for $68$) wich besides could lead to a two digit number such as 237 with its factor $3$.


Arthur Benjamin wrote a paper called "Squaring, Cubing, and Cube Rooting" which contains a great shortcut for cubing. It's similar to the method posted by callculus, but the math is broken up in a different way.

Let's refer to the number to be cubed as $x$. The closest multiple of 10 to the given number will be called $z$, and we'll define the difference, $d$, as $x-z$. Note that, since $z$ is the closest multiple of 10, $d$ will always range from $-5$ to $5$. The following formula is worked from the inside out:


A couple of things to remember that help make things quicker: a) $3d^{2}$ will always be positive, and b) Since d can only be $\pm1, \pm2, \pm3, \pm4,$ or $\pm5,$ then $3d^{2}$ can only ever be $3, 12, 27, 48,$ or $75$ respectively.

Let's try this with your example, $68$. So, $x=68, z=70,$ and $d=-2$.

Start with the $3d^{2}$. You need to get to the point where you know the answers from memory. Since we're dealing with $d=-2$, this is equal to 12.

Multiply this by $z$:


Next, add $d^{3}$ to this total. Since $d=-2$, then $d^{3}=-8$:


After this, there's just 2 more steps, and they won't affect the last 2 digits, so you can just write down the last two digit ($32$ in this case), and remember only the remaining digits ($8$ in this this case).

The net step is to multiply $(z+3d)\times(z^{2})$. Since $z=70$ and $d=-2$, then $70+(-6)=64$. To make things easier, just multiply that by $(\frac{1}{10}z)^{2}$. In this case, you'd be multiplying by 49 instead of 4,900. So, for this step:

$$64(49) \\ 64(50-1) \\ 3,200-64 \\ 3,136$$

Finally, recall the 8 (or whatever digits they happen to be) from earlier. Simply add that amount to this total:


Now, write this number down to the left of the digits you wrote down earlier (the $32$ in this case), and you have your answer!


Practice this by cubing smaller numbers to get used to the pattern, and then work your way up to higher numbers as you get more comfortable with the process.

This works well for 2-digit numbers. It can be used for 3-digit numbers, as well, by defining $z$ as the nearest multiple of 100, but you obviously need to be comfortable with quickly squaring and cubing 2-digit numbers first.

After linking to the above article, Colin Beveridge of http://www.flyingcoloursmaths.co.uk/ shared the following shortcut with me, used for cubing 2-digit numbers ending in $5$.

Given a number of the form $10n+5$, the thousands digits can be calculated as:

$$((n(n+1)(2n+1))/2)+\left \lfloor n/4 \right \rfloor$$

That $\left \lfloor n/4 \right \rfloor$ represents the floor function (divide, and always round down to the nearest whole number).

The last 3 digits cycle in a pattern:

n:                   0     1     2     3     4     5     6     7     8     9

(10n+5)^3 mod 1000: 125   375   625   875   125   375   625   875   125   375

If you prefer not to memorize this table, you can work out the following formula to get the same result:

$$125((2(n \ mod \ 4))+1)$$

For example, what is 65^3?

In this case, n=6. Start by calculating $((n(n+1)(2n+1))/2)$, keeping in mind that either $n$ or $n+1$ will always be even:

$$((6)(7)(13))/2 \\ (3)(7)(13) \\ (21)(13) \\ 273$$

Work out $\left \lfloor n/4 \right \rfloor$ for n=6:

$$\left \lfloor 6/4 \right \rfloor \\ \left \lfloor 1.5 \right \rfloor \\ 1$$

Add those 2 together:


Finally, because n=6, work out by memorization or calculation that this means you should add 625 to the end. The solution, then, is:



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.