# Cube (algebra)

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In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 23 = 8 or (x + 1)3.

## Contents

The cube is also the number multiplied by its square:

n3 = n × n2 = n × n × n.

The cube function is the function xx3 (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 perfect cubes up to 603 are (sequence 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 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:

 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, ${\displaystyle 6=2^{3}+(-1)^{3}+(-1)^{3}}$. (note that the integer 0 cannot be written in this way, by Fermat’s Last Theorem) Integers congruent to ±4 modulo 9 are excluded because they cannot be written as the sum of three cubes, since cubes are congruent to 0 or ±1 mod 9.

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] [3]

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

One solution to ${\displaystyle x^{3}+y^{3}+z^{3}=n}$ is given in the table below for n ≤ 160, 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 ${\displaystyle k^{3}+(-k)^{3}+n^{3}=n^{3}}$ or ${\displaystyle (n+6k^{3}n)^{3}+(n-6k^{3}n)^{3}+(-6k^{2}n)^{3}=2n^{3}}$ (since they are infinite families of solutions, however, for n3, ALL solutions are of the form a=q[1−(x−3y)(x^2+3y^2)], b=−q[1−(x+3y)(x^2+3y^2)], c=q[(x^2 +3y^2)^2−(x+3y)], n=q[(x^2+3y^2)^2−(x−3y)], [4] i.e. all solutions are in infinitely families, there is no solution which is not in infinitely families, but for 2n3, the only infinity family is ${\displaystyle (n+6k^{3}n)^{3}+(n-6k^{3}n)^{3}+(-6k^{2}n)^{3}=2n^{3}}$, i.e. all solutions with none of x+y, x+z, y+z, is 2n, are not in infinitely families, numbers of the form n3 or 2n3 are the only numbers with representations that can be parameterized by quartic polynomials), satisfies 0 ≤ |x||y||z|, and has minimal values for |z| and |y| (tested in this order). [5] [6] [7]

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.

The only remaining unsolved cases up to 1000 are 114, 390, 627, 633, 732, 921, and 975, and there are no known primitive solutions (i.e. ${\displaystyle \gcd(x,y,z)=1}$) for 192, 375, and 600 [8] [9]

Note that there is only one known solution for 3 besides (1, 1, 1) and (−4, −4, 5):

${\displaystyle 3=569936821221962380720^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}}$
 n x y z n x y z n x y z n x y z 1 9 10 −12 42 12602123297335631 80435758145817515 −80538738812075974 81 10 17 −18 120 946 1531 −1643 2 1214928 3480205 −3528875 43 2 2 3 82 −11 −11 14 123 −1 −1 5 3 1 1 1 44 −5 −7 8 83 −2 3 4 124 0 −1 5 6 −1 −1 2 45 2 −3 4 84 −8241191 −41531726 41639611 125 −3 −4 6 7 0 −1 2 46 −2 3 3 87 −1972 −4126 4271 126 0 1 5 8 9 15 −16 47 6 7 −8 88 3 −4 5 127 −1 4 4 9 0 1 2 48 −23 −26 31 89 6 6 −7 128 553 1152 −1193 10 1 1 2 51 602 659 −796 90 −1 3 4 129 1 4 4 11 −2 −2 3 52 23961292454 60702901317 −61922712865 91 0 3 4 132 −1 2 5 12 7 10 −11 53 −1 3 3 92 1 3 4 133 0 2 5 15 −1 2 2 54 −7 −11 12 93 −5 −5 7 134 1 2 5 16 −511 −1609 1626 55 1 3 3 96 10853 13139 −15250 135 2 −6 7 17 1 2 2 56 −11 −21 22 97 −1 −3 5 136 225 582 −593 18 −1 −2 3 57 1 −2 4 98 0 −3 5 137 −9 −11 13 19 0 −2 3 60 −1 −4 5 99 2 3 4 138 −77 −86 103 20 1 −2 3 61 0 −4 5 100 −3 −6 7 141 2 2 5 21 −11 −14 16 62 2 3 3 101 −3 4 4 142 −3 −7 8 24 −2901096694 −15550555555 15584139827 63 0 −1 4 102 118 229 −239 143 7023 84942 −84958 25 −1 −1 3 64 −3 −5 6 105 −4 −7 8 144 −2 3 5 26 0 −1 3 65 0 1 4 106 2 −3 5 145 −7 −8 10 27 −4 −5 6 66 1 1 4 107 −28 −48 51 146 −5 −9 10 28 0 1 3 69 2 −4 5 108 −948 −1165 1345 147 −50 −56 67 29 1 1 3 70 11 20 −21 109 −2 −2 5 150 260 317 −367 30 −283059965 −2218888517 2220422932 71 −1 2 4 110 109938919 16540290030 −16540291649 151 −1 3 5 33 −2736111468807040 −8778405442862239 8866128975287528 72 7 9 −10 111 −296 −881 892 152 0 3 5 34 −1 2 3 73 1 2 4 114 ? ? ? 153 1 3 5 35 0 2 3 74 66229832190556 283450105697727 −284650292555885 115 −6 −10 11 154 −4 −5 7 36 1 2 3 75 4381159 435203083 −435203231 116 −1 −2 5 155 3 4 4 37 0 −3 4 78 26 53 −55 117 0 −2 5 156 68844645625 2232194323 −68845427846 38 1 −3 4 79 −19 −33 35 118 3 3 4 159 80 119 −130 39 117367 134476 −159380 80 69241 103532 −112969 119 −2 −6 7 160 2 3 5

### 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. [10]

Both of these statements are also true for the equation [11] x3 + y3 = nz3 for n = 3, 4, 5, 10, 11, 14, 18, 21, 23, 24, 25, 29, 32, 36, 38, 39, 40, 41, 44, 45, 46, 47, 52, 55, 57, 59, 60, 66, 73, 74, 76, 77, 80, 81, 82, 83, 88, 93, 95, 99, 100, ... (sequence in the OEIS ), also for n = k3 and 2k3, there are no nontrivial solutions, i.e. solutions other than 03 + (kx)3 = (k3)x3 and (kx)3 + (kx)3 = (2k3)x3, for all other positive integers values of k, there are infinitely many primitive solutions, see next section.

### 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. CharlesWheatstone  ( 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,

${\displaystyle 1^{3}+3^{3}+\dots +9^{3}=(7\cdot 5)^{2}}$
${\displaystyle 1^{3}+3^{3}+\dots +57^{3}=(41\cdot 29)^{2}}$
${\displaystyle 28=2^{2}(2^{3}-1)=1^{3}+3^{3}}$
${\displaystyle 496=2^{4}(2^{5}-1)=1^{3}+3^{3}+5^{3}+7^{3}}$
${\displaystyle 8128=2^{6}(2^{7}-1)=1^{3}+3^{3}+5^{3}+7^{3}+9^{3}+11^{3}+13^{3}+15^{3}}$

### 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. [12]

### 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, [13] and there are rationals that are not the sum of two rational cubes. [14]

However, there are rational numbers that are not the sum of two rational cubes. [15] For positive integers, the positive integers that is the sum of two positive noninteger rational cubes (also the positive integer that can be written as sum of two rational cubes in infinitely many different ways) are

6, 7, 9, 12, 13, 15, 17, 19, 20, 22, 26, 28, 30, 31, 33, 34, 35, 37, 42, 43, 48, 49, 50, 51, 53, 56, 58, 61, 62, 63, 65, 67, 68, 69, 70, 71, 72, 75, 78, 79, 84, 85, 86, 87, 89, 90, 91, 92, 94, 96, 97, 98, 103, 104, 105, 106, 107, 110, 114, 115, 117, 120, 123, 124, 126, 127, 130, 132, 133, 134, 136, 139, 140, 141, 142, 143, 151, 152, 153, 156, 157, 159, 160, ... (sequence in the OEIS ) (also see and )

The following is a list of the integer solution to ${\displaystyle \left({\frac {a}{c}}\right)^{3}+\left({\frac {b}{c}}\right)^{3}=n}$ (i.e. ${\displaystyle a^{3}+b^{3}=nc^{3}}$) with the smallest c > 1. (we let a < b) [16] [17] [18] (For the sequence of the solutions of a and b, see and , and for the same sequences of allowing negative a-values, see , and for the a, b and c)

 n a b c 6 17 37 21 7 4 5 3 9 415280564497 676702467503 348671682660 12 19 89 39 13 2 7 3 15 397 683 294 17 ? ? ? 19 3 5 2 20 1 19 7 22 17299 25469 9954 26 53 75 28 28 ? ? ? 30 107 163 57 31 ? ? ? 33 523 1853 582 34 ? ? ? 35 ? ? ? 37 18 19 7 42 ? ? ? 43 1 7 2 48 34 74 21 49 5308 14465 4017 50 ? ? ?

## 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, i 3 = −i.

The derivative of x3 equals 3x2.

Cubes occasionally have the surjective property in other fields, such as in for such prime p that p ≠ 1 (mod 3), [19] 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). [20] [21] Cubic equations were known to the ancient Greek mathematician Diophantus. [22] Hero of Alexandria devised a method for calculating cube roots in the 1st century CE. [23] 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. [24]

## Related Research Articles

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