Cube (algebra)

The cube is also the number multiplied by its square:

n³ = n × n² = n × n × n.

The cube function is the function x ↦ x³ (often denoted y = x³) that maps a number to its cube. It is an odd function, as

(−n)³ = −(n³).

The volume of a geometric cube is the cube of its side length, giving rise to the name. The inverse operation that consists of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power.

The graph of the cube function is known as the cubic parabola. Because the cube function is an odd function, this curve has a center of symmetry at the origin, but no axis of symmetry.

In integers

A cube number, or a perfect cube, or sometimes just a cube, is a number which is the cube of an integer. The non-negative perfect cubes up to 60³ are (sequence A000578 in the OEIS):

Geometrically speaking, a positive integer m is a perfect cube if and only if one can arrange m solid unit cubes into a larger, solid cube. For example, 27 small cubes can be arranged into one larger one with the appearance of a Rubik's Cube, since 3 × 3 × 3 = 27.

The difference between the cubes of consecutive integers can be expressed as follows:

n³ − (n − 1)³ = 3(n − 1)n + 1.

or

(n + 1)³ − n³ = 3(n + 1)n + 1.

There is no minimum perfect cube, since the cube of a negative integer is negative. For example, (−4) × (−4) × (−4) = −64.

Base ten

Unlike perfect squares, perfect cubes do not have a small number of possibilities for the last two digits. Except for cubes divisible by 5, where only 25, 75 and 00 can be the last two digits, any pair of digits with the last digit odd can occur as the last digits of a perfect cube. With even cubes, there is considerable restriction, for only 00, o2, e4, o6 and e8 can be the last two digits of a perfect cube (where o stands for any odd digit and e for any even digit). Some cube numbers are also square numbers; for example, 64 is a square number (8 × 8) and a cube number (4 × 4 × 4). This happens if and only if the number is a perfect sixth power (in this case 2⁶).

The last digits of each 3rd power are:

It is, however, easy to show that most numbers are not perfect cubes because all perfect cubes must have digital root 1, 8 or 9. That is their values modulo 9 may be only 0, 1, and 8. Moreover, the digital root of any number's cube can be determined by the remainder the number gives when divided by 3:

• If the number x is divisible by 3, its cube has digital root 9; that is,

  ${\displaystyle \text{if}x\equiv 0(\mathrm{mod}3)\text{then}{x}^3\equiv 0(\mathrm{mod}9)\text{ (actually}0(\mathrm{mod}27)\text{)};}$

• If it has a remainder of 1 when divided by 3, its cube has digital root 1; that is,

  ${\displaystyle \text{if}x\equiv 1(\mathrm{mod}3)\text{then}{x}^3\equiv 1(\mathrm{mod}9);}$

• If it has a remainder of 2 when divided by 3, its cube has digital root 8; that is,

  ${\displaystyle \text{if}x\equiv 2(\mathrm{mod}3)\text{then}{x}^3\equiv 8(\mathrm{mod}9).}$

Sums of two cubes

Sums of three cubes

It is conjectured that every integer (positive or negative) not congruent to ±4 modulo 9 can be written as a sum of three (positive or negative) cubes in infinitely many ways. For example, ${\displaystyle 6=2^3+(-1{)}^3+(-1{)}^3}$. Integers congruent to ±4 modulo 9 are excluded because they cannot be written as the sum of three cubes.

The smallest such integer for which such a sum is not known is 114. In September 2019, the previous smallest such integer with no known 3-cube sum, 42, was found to satisfy this equation:

${\displaystyle 42=(-80538738812075974{)}^3+{80435758145817515}^3+{12602123297335631}^3.}$

One solution to ${\displaystyle x^3+y^3+z^3=n}$ is given in the table below for n ≤ 78, and n not congruent to 4 or 5 modulo 9. The selected solution is the one that is primitive (gcd(x, y, z) = 1), is not of the form ${\displaystyle c^3+(-c{)}^3+n^3=n^3}$ or ${\displaystyle (n+6n{c}^3{)}^3+(n-6n{c}^3{)}^3+(-6n{c}^2{)}^3=2n^3}$ (since they are infinite families of solutions), satisfies 0 ≤ |x| ≤ |y| ≤ |z|, and has minimal values for |z| and |y| (tested in this order).

Only primitive solutions are selected since the non-primitive ones can be trivially deduced from solutions for a smaller value of n. For example, for n = 24, the solution ${\displaystyle 2^3+2^3+2^3=24}$ results from the solution ${\displaystyle 1^3+1^3+1^3=3}$ by multiplying everything by ${\displaystyle 8=2^3.}$ Therefore, this is another solution that is selected. Similarly, for n = 48, the solution (x, y, z) = (−2, −2, 4) is excluded, and this is the solution (x, y, z) = (−23, −26, 31) that is selected.

Fermat's Last Theorem for cubes

The equation x³ + y³ = z³ has no non-trivial (i.e. xyz ≠ 0) solutions in integers. In fact, it has none in Eisenstein integers.

Both of these statements are also true for the equation x³ + y³ = 3z³.

Sum of first n cubes

The sum of the first n cubes is the nth triangle number squared:

${\displaystyle 1^3+2^3+\cdots +n^3=(1+2+\cdots +n{)}^2={\left(\frac{n(n+1)}{2}\right)}^2.}$

Proofs. Charles Wheatstone (1854) gives a particularly simple derivation, by expanding each cube in the sum into a set of consecutive odd numbers. He begins by giving the identity

${\displaystyle n^3=\underset{n\text{ consecutive odd numbers}}{\underbrace{\left(n^2-n+1\right)+\left(n^2-n+1+2\right)+\left(n^2-n+1+4\right)+\cdots +\left(n^2+n-1\right)}}.}$

That identity is related to triangular numbers ${\displaystyle T_n}$ in the following way:

${\displaystyle n^3=\sum _{k=T_{n-1}+1}^{T_n}(2k-1),}$

and thus the summands forming ${\displaystyle n^3}$ start off just after those forming all previous values ${\displaystyle 1^3}$ up to ${\displaystyle (n-1{)}^3}$. Applying this property, along with another well-known identity:

${\displaystyle n^2=\sum _{k=1}^n(2k-1),}$

we obtain the following derivation:

${\displaystyle \begin{array}{cc}\sum _{k=1}^n{k}^3& =1+8+27+64+\cdots +n^3\\ & =\underset{1^3}{\underbrace{1}}+\underset{2^3}{\underbrace{3+5}}+\underset{3^3}{\underbrace{7+9+11}}+\underset{4^3}{\underbrace{13+15+17+19}}+\cdots +\underset{n^3}{\underbrace{\left(n^2-n+1\right)+\cdots +\left(n^2+n-1\right)}}\\ & =\underset{{\left(\frac{n^2+n}{2}\right)}^2}{\underbrace{\underset{3^2}{\underbrace{\underset{2^2}{\underbrace{\underset{1^2}{\underbrace{1}}+3}}+5}}+\cdots +\left(n^2+n-1\right)}}\\ & =(1+2+\cdots +n{)}^2\\ & =(\sum _{k=1}^{n}k{)}^2.\end{array}}$

In the more recent mathematical literature, Stein (1971) uses the rectangle-counting interpretation of these numbers to form a geometric proof of the identity (see also Benjamin, Quinn & Wurtz 2006); he observes that it may also be proved easily (but uninformatively) by induction, and states that Toeplitz (1963) provides "an interesting old Arabic proof". Kanim (2004) provides a purely visual proof, Benjamin & Orrison (2002) provide two additional proofs, and Nelsen (1993) gives seven geometric proofs.

For example, the sum of the first 5 cubes is the square of the 5th triangular number,

${\displaystyle 1^3+2^3+3^3+4^3+5^3={15}^2}$

A similar result can be given for the sum of the first y odd cubes,

${\displaystyle 1^3+3^3+\cdots +(2y-1{)}^3=(xy{)}^2}$

but x, y must satisfy the negative Pell equation x² − 2y² = −1. For example, for y = 5 and 29, then,

${\displaystyle 1^3+3^3+\cdots +9^3=(7\cdot 5{)}^2}$

${\displaystyle 1^3+3^3+\cdots +{57}^3=(41\cdot 29{)}^2}$

and so on. Also, every even perfect number, except the lowest, is the sum of the first 2^{⁠p−1/2⁠}
odd cubes (p = 3, 5, 7, ...):

${\displaystyle 28=2^2(2^3-1)=1^3+3^3}$

${\displaystyle 496=2^4(2^5-1)=1^3+3^3+5^3+7^3}$

${\displaystyle 8128=2^6(2^7-1)=1^3+3^3+5^3+7^3+9^3+{11}^3+{13}^3+{15}^3}$

Sum of cubes of numbers in arithmetic progression

There are examples of cubes of numbers in arithmetic progression whose sum is a cube:

${\displaystyle 3^3+4^3+5^3=6^3}$

${\displaystyle {11}^3+{12}^3+{13}^3+{14}^3={20}^3}$

${\displaystyle {31}^3+{33}^3+{35}^3+{37}^3+{39}^3+{41}^3={66}^3}$

with the first one sometimes identified as the mysterious Plato's number. The formula F for finding the sum of n cubes of numbers in arithmetic progression with common difference d and initial cube a³,

${\displaystyle F(d,a,n)=a^3+(a+d{)}^3+(a+2d{)}^3+\cdots +(a+dn-d{)}^3}$

is given by

${\displaystyle F(d,a,n)=(n/4)(2a-d+dn)(2a^2-2ad+2adn-d^{2}n+d^2{n}^2)}$

A parametric solution to

${\displaystyle F(d,a,n)=y^3}$

is known for the special case of d = 1, or consecutive cubes, as found by Pagliani in 1829.

Cubes as sums of successive odd integers

In the sequence of odd integers 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, ..., the first one is a cube (1 = 1³); the sum of the next two is the next cube (3 + 5 = 2³); the sum of the next three is the next cube (7 + 9 + 11 = 3³); and so forth.

Waring's problem for cubes

Every positive integer can be written as the sum of nine (or fewer) positive cubes. This upper limit of nine cubes cannot be reduced because, for example, 23 cannot be written as the sum of fewer than nine positive cubes:

23 = 2³ + 2³ + 1³ + 1³ + 1³ + 1³ + 1³ + 1³ + 1³.