1.3 Cubic Numbers

You know the cube from geometry: a solid with equal sides meeting at right angles. How many unit cubes (side 1 cm) pack inside a larger cube?

The numbers 1, 8, 27, 64, 125, \dots are the perfect cubes, because each is a number multiplied by itself three times:

1 = 1\times1\times1,\quad 8 = 2\times2\times2,\quad 27 = 3\times3\times3,\ \dots

We record this with the cubing notation:

n\times n\times n = n^3, \qquad \text{read as } ``n \text{ cubed}".

As with squares, 10 is not a cube, it sits between 2^3 = 8 and 3^3 = 27, and so does every number from 9 to 26. Fractions, decimals and negatives can be cubed too:

\left(\tfrac{2}{3}\right)^3 = \tfrac{8}{27},\qquad (1.2)^3 = 1.728,\qquad (-5)^3 = -125.

Math Talk For squares, the only possible last digits were 0,1,4,5,6,9. Build a table of cubes from 1^3 to 10^3 and inspect their last digits. What is different this time? (You should discover that cubes can end in every digit from 0 to 9.) And can a cube ever end in exactly two zeros, like \dots00?

A guessing trick: the last digit of a cube

Cubes hide a neat secret in their final digit. For most digits the cube ends in the same digit (1\to1, 4\to4, 5\to5, 6\to6, 9\to9, 0\to0), while 2 and 8 swap (2^3 = 8, 8^3 = 512) and 3 and 7 swap (3^3 = 27, 7^3 = 343). This lets you guess a cube root’s last digit instantly. For example, a perfect cube ending in 3 must have a root ending in 7. We will put this to work below.

The most famous taxicab in mathematics

Did You Know: The Hardy–Ramanujan Number While the brilliant Indian mathematician Srinivasa Ramanujan lay ill in a hospital near Cambridge, his collaborator G. H. Hardy came by and mentioned, in passing, that his taxi bore the “rather dull” number 1729. Ramanujan disagreed instantly: it is the smallest number that can be written as a sum of two cubes in two different ways.

1729 = 1^3 + 12^3 = 9^3 + 10^3.

Such numbers are now called taxicab numbers. Ramanujan’s friend John Littlewood later said that for him, “every positive integer was a personal friend.”

Try This The next two taxicab numbers after 1729 are 4104 and 13832. Find the two different ways each can be expressed as a sum of two positive cubes. (Hint: for 4104, one of the pairs uses 2^3.)

Cubes and consecutive odd numbers

Odd numbers play a role with cubes too:

\begin{aligned} 1 &= 1 = 1^3\\ 3+5 &= 8 = 2^3\\ 7+9+11 &= 27 = 3^3\\ 13+15+17+19 &= 64 = 4^3\\ 21+23+25+27+29 &= 125 = 5^3. \end{aligned}

The nth cube is the sum of n consecutive odd numbers. So the block 91+93+95+\dots+109 (ten consecutive odd numbers) must add to 10^3 = 1000, without any adding at all!

Cube roots

Since 8 = 2^3, we call 2 the cube root of 8 and write \sqrt[3]{8} = 2. In general:

\text{If } y = x^3, \text{ then } x = \sqrt[3]{y}, \qquad\text{and } \sqrt[3]{n^3} = n.

So \sqrt[3]{27} = 3 and \sqrt[3]{1000} = 10. To check whether a number is a perfect cube, return to prime factorisation, but now the factors must split into three identical groups.

Worked Example 5

Is 9261 a perfect cube?

9261 = 3\times 3\times 3\times 7\times 7\times 7 = (3\times 7)\times(3\times 7)\times(3\times 7) = (3\times 7)^3 = 21^3.

Three identical groups, so \sqrt[3]{9261} = 21.

Is 600 a perfect cube? 600 = 2\times 2\times 2\times 3\times 5\times 5. The single 3 and the pair of 5s cannot be packed into three identical groups, so 600 is not a perfect cube.

Notice the pattern in the table below: each prime factor of a number appears three times in the prime factorisation of its cube.

Number Its cube
6 = 2\times3 6^3 = 216 = 2^3\times 3^3
10 = 2\times5 10^3 = 1000 = 2^3\times 5^3
14 = 2\times7 14^3 = 2744 = 2^3\times 7^3
20 = 2\times2\times5 20^3 = 8000 = 2^3\times 2^3\times 5^3

A quick self-check: \sqrt[3]{216} = 6,\quad \sqrt[3]{343} = 7,\quad \sqrt[3]{512} = 8.

Successive differences

For perfect squares, taking differences once gives the odd numbers, and a second round gives a constant 2:

\underbrace{1,\ 4,\ 9,\ 16,\ 25,\ 36}_{\text{squares}}\;\to\;\underbrace{3,\ 5,\ 7,\ 9,\ 11}_{\text{level 1}}\;\to\;\underbrace{2,\ 2,\ 2,\ 2}_{\text{level 2}}.

Try This Do the same with the perfect cubes 1, 8, 27, 64, 125, 216, \dots Take successive differences, level by level, until every value at a level is the same. How many levels does it take, and what is that final constant? (You should find it settles after three levels.)