# Cube (algebra)

In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice:

n3 = n × n × n.

It is also the number multiplied by its square:

n3 = n × n2.

This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation 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.

Both cube and cube root are odd functions:

(−n)3 = −(n3).

The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 23 = 8 or (x + 1)3.

The graph of the cube function f: xx3 (or the equation y = x3) is known as the cubic parabola. Because cube is an odd function, this curve has a point of symmetry in 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 perfect cubes up to 603 are (sequence A000578 in the OEIS):

 03 = 0 13 = 1 113 = 1331 213 = 9261 313 = 29,791 413 = 68,921 513 = 132,651 23 = 8 123 = 1728 223 = 10,648 323 = 32,768 423 = 74,088 523 = 140,608 33 = 27 133 = 2197 233 = 12,167 333 = 35,937 433 = 79,507 533 = 148,877 43 = 64 143 = 2744 243 = 13,824 343 = 39,304 443 = 85,184 543 = 157,464 53 = 125 153 = 3375 253 = 15,625 353 = 42,875 453 = 91,125 553 = 166,375 63 = 216 163 = 4096 263 = 17,576 363 = 46,656 463 = 97,336 563 = 175,616 73 = 343 173 = 4913 273 = 19,683 373 = 50,653 473 = 103,823 573 = 185,193 83 = 512 183 = 5832 283 = 21,952 383 = 54,872 483 = 110,592 583 = 195,112 93 = 729 193 = 6859 293 = 24,389 393 = 59,319 493 = 117,649 593 = 205,379 103 = 1000 203 = 8000 303 = 27,000 403 = 64,000 503 = 125,000 603 = 216,000

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)3n3 = 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 be 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:

 0 1 8 7 4 5 6 3 2 9

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 −1, 1 and 0. 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,
${\displaystyle {\text{if}}\quad x\equiv 0{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv 0{\pmod {9}}{\text{ (actually}}\quad 0{\pmod {27}}{\text{)}};}$
• If it has a remainder of 1 when divided by 3, its cube has digital root 1; that is,
${\displaystyle {\text{if}}\quad x\equiv 1{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv 1{\pmod {9}};}$
• If it has a remainder of 2 when divided by 3, its cube has digital root 8; that is,
${\displaystyle {\text{if}}\quad x\equiv -1{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv -1{\pmod {9}}.}$

### 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.

### 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 with infinitely many ways,[1] for example, ${\displaystyle 6=2^{3}+(-1)^{3}+(-1)^{3}}$. (Integers congruent to ±4 modulo 9 cannot be so written.) 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: [2]

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

The solutions to ${\displaystyle x^{3}+y^{3}+z^{3}=n}$ for n ≤ 78 (where n is not = 4 or 5 mod 9) with smallest |z| and smallest |y|, 0 ≤ |x| ≤ |y| ≤ |z|, gcd(x,y,z) = 1 (i.e. only primitive solutions are selected), and none of x+y, y+z, z+x is 0 (if no this condition, then for every positive cube z3, all (x, −x, z) are solutions, and these solutions are all trivial) are given below:[3][4][5] (We only select "primitive solutions", i.e. gcd(x,y,z) must be 1, e.g. for n=24, the solution (x,y,z) = (2, 2, 2) is not allowed, since its gcd is 2 (not 1), thus we select (x,y,z) = (-2901096694, -15550555555, 15584139827), another example is for n=48, the solution (x,y,z) = (-2, -2, 4) is also not allowed, and we select (x,y,z) = (-23, -26, 31)) (sequences A060465, A060466 and A060467 in OEIS)

 n x y z n x y z 1 9 10 −12 39 117367 134476 −159380 2 0 1 1 42 12602123297335631 80435758145817515 −80538738812075974 3 1 1 1 43 2 2 3 6 −1 −1 2 44 −5 −7 8 7 0 −1 2 45 2 −3 4 8 9 15 −16 46 −2 3 3 9 0 1 2 47 6 7 −8 10 1 1 2 48 −23 −26 31 11 −2 −2 3 51 602 659 −796 12 7 10 −11 52 23961292454 60702901317 −61922712865 15 −1 2 2 53 −1 3 3 16 −511 −1609 1626 54 −7 −11 12 17 1 2 2 55 1 3 3 18 −1 −2 3 56 −11 −21 22 19 0 −2 3 57 1 −2 4 20 1 −2 3 60 −1 −4 5 21 −11 −14 16 61 0 −4 5 24 −2901096694 −15550555555 15584139827 62 2 3 3 25 −1 −1 3 63 0 −1 4 26 0 −1 3 64 −3 −5 6 27 −4 −5 6 65 0 1 4 28 0 1 3 66 1 1 4 29 1 1 3 69 2 −4 5 30 −283059965 −2218888517 2220422932 70 11 20 −21 33 −2736111468807040 −8778405442862239 8866128975287528 71 −1 2 4 34 −1 2 3 72 7 9 −10 35 0 2 3 73 1 2 4 36 1 2 3 74 66229832190556 283450105697727 −284650292555885 37 0 −3 4 75 4381159 435203083 −435203231 38 1 −3 4 78 26 53 −55

### 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.[6]

Both of these statements are also true for the equation[7] x3 + y3 = 3z3.

### Sum of first n cubes

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

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

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

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

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

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,

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

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

${\displaystyle 1^{3}+3^{3}+\dots +9^{3}=(7\cdot 5)^{2}}$
${\displaystyle 1^{3}+3^{3}+\dots +57^{3}=(41\cdot 29)^{2}}$

and so on. Also, every even perfect number, except the lowest, is the sum of the first 2p−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, but only sporadic solutions are known for integer d > 1, such as d = 2, 3, 5, 7, 11, 13, 37, 39, etc.[8]

### 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.

## In rational numbers

Every positive rational number is the sum of three positive rational cubes,[9] and there are rationals that are not the sum of two rational cubes.[10]

## In real numbers, other fields, and rings

In real numbers, the cube function preserves the order: larger numbers have larger cubes. In other words, cubes (strictly) monotonically increase. Also, its codomain is the entire real line: the function xx3 : RR is a surjection (takes all possible values). Only three numbers are equal to their own cubes: −1, 0, and 1. If −1 < x < 0 or 1 < x, then x3 > x. If x < −1 or 0 < x < 1, then x3 < x. All aforementioned properties pertain also to any higher odd power (x5, x7, …) of real numbers. Equalities and inequalities are also true in any ordered ring.

Volumes of similar Euclidean solids are related as cubes of their linear sizes.

In complex numbers, the cube of a purely imaginary number is also purely imaginary. For example, i3 = −i.

The derivative of x3 equals 3x2.

Cubes occasionally have the surjective property in other fields, such as in Fp for such prime p that p ≠ 1 (mod 3),[11] but not necessarily: see the counterexample with rationals above. Also in F7 only three elements 0, ±1 are perfect cubes, of seven total. −1, 0, and 1 are perfect cubes anywhere and the only elements of a field equal to the own cubes: x3x = x(x − 1)(x + 1).

## History

Determination of the cubes of large numbers was very common in many ancient civilizations. Mesopotamian mathematicians created cuneiform tablets with tables for calculating cubes and cube roots by the Old Babylonian period (20th to 16th centuries BC).[12][13] Cubic equations were known to the ancient Greek mathematician Diophantus.[14] Hero of Alexandria devised a method for calculating cube roots in the 1st century CE.[15] Methods for solving cubic equations and extracting cube roots appear in The Nine Chapters on the Mathematical Art, a Chinese mathematical text compiled around the 2nd century BCE and commented on by Liu Hui in the 3rd century CE.[16] The Indian mathematician Aryabhata wrote an explanation of cubes in his work Aryabhatiya. In 2010 Alberto Zanoni found a new algorithm[17] to compute the cube of a long integer in a certain range, faster than squaring-and-multiplying.

## Notes

1. Huisman, Sander G. (27 Apr 2016). "Newer sums of three cubes". arXiv:1604.07746 [math.NT].
2. "NEWS: The Mystery of 42 is Solved - Numberphile" https://www.youtube.com/watch?v=zyG8Vlw5aAw
3. List of solutions of x^3 + y^3 + z^3 = k for k < 1000
5. Three cubes
6. Hardy & Wright, Thm. 227
7. Hardy & Wright, Thm. 232
8. "A Collection of Algebraic Identities".
9. Hardy & Wright, Thm. 234
10. Hardy & Wright, Thm. 233
11. The multiplicative group of Fp is cyclic of order p − 1, and if it is not divisible by 3, then cubes define a group automorphism.
12. Cooke, Roger (8 November 2012). The History of Mathematics. John Wiley & Sons. p. 63. ISBN 978-1-118-46029-0.
13. Nemet-Nejat, Karen Rhea (1998). Daily Life in Ancient Mesopotamia. Greenwood Publishing Group. p. 306. ISBN 978-0-313-29497-6.
14. Van der Waerden, Geometry and Algebra of Ancient Civilizations, chapter 4, Zurich 1983 ISBN 0-387-12159-5
15. Smyly, J. Gilbart (1920). "Heron's Formula for Cube Root". Hermathena. Trinity College Dublin. 19 (42): 64–67. JSTOR 23037103.
16. Crossley, John; W.-C. Lun, Anthony (1999). The Nine Chapters on the Mathematical Art: Companion and Commentary. Oxford University Press. pp. 176, 213. ISBN 978-0-19-853936-0.
17. Bodrato, Marco; Zanoni, Alberto (2012). "A New Algorithm for Long Integer Cube Computation with Some Insight into Higher Powers". Lecture Notes in Computer Science: 34–46. doi:10.1007/978-3-642-32973-9_4.