Cube (algebra)

"Third power" redirects here. For the band, see Third Power.

"Cubed" redirects here. For other uses, see Cube (disambiguation).

"³" redirects here. Its literal meaning is the numeral "3" in superscript.

y = x³ for values of 1 ≤ x ≤ 25.

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.

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 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)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:

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,

  ${\text{if}}\quad x\equiv 0{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv 0{\pmod {9}}{\text{ (actually}}\quad 0{\pmod {27}}{\text{)}};$

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

  ${\text{if}}\quad x\equiv 1{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv 1{\pmod {9}};$

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

  ${\text{if}}\quad x\equiv -1{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv -1{\pmod {9}}.$

Waring's problem for cubes

Main article: Waring's problem

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

Main article: 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. For example, $6=2^{3}+(-1)^{3}+(-1)^{3}$. (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:

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

One solution to $x^{3}+y^{3}+z^{3}=n$ 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 $k^{3}+(-k)^{3}+n^{3}=n^{3}$ or $(n+6k^{3}n)^{3}+(n-6k^{3}n)^{3}+(-6k^{2}n)^{3}=2n^{3}$ (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)], 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 $(n+6k^{3}n)^{3}+(n-6k^{3}n)^{3}+(-6k^{2}n)^{3}=2n^{3}$, 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).

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 $2^{3}+2^{3}+2^{3}=24$ results from the solution $1^{3}+1^{3}+1^{3}=3$ by multiplying everything by $8=2^{3}.$ 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. $\gcd(x,y,z)=1$) for 192, 375, and 600

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

$3=569936821221962380720^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}$

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

Main article: Fermat's Last Theorem

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 = 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 A185345 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:

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

Visual proof that 1³ + 2³ + 3³ + 4³ + 5³ = (1 + 2 + 3 + 4 + 5)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

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

That identity is related to triangular numbers $T_{n}$ in the following way:

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

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

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

we obtain the following derivation:

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

Visual demonstration that the square of a triangular number equals a sum of cubes.

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,

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

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

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

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

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

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

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

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

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

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

Sum of cubes of numbers in arithmetic progression

One interpretation of Plato's number, 3³ + 4³ + 5³ = 6³

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

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

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

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

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,

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

is given by

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

A parametric solution to

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

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.

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, and there are rationals that are not the sum of two rational cubes.

However, there are rational numbers that are not the sum of two rational cubes. 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 A228499 in the OEIS) (also see OEIS: A159843 and OEIS: A020898)

The following is a list of the integer solution to $\left({\frac {a}{c}}\right)^{3}+\left({\frac {b}{c}}\right)^{3}=n$ (i.e. $a^{3}+b^{3}=nc^{3}$) with the smallest c > 1. (we let a < b) (For the sequence of the solutions of a and b, see OEIS: A254326 and OEIS: A254324, and for the same sequences of allowing negative a-values, see OEIS: A190580, OEIS: A190356 and OEIS: A190581 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

Further information: cubic function

y = x³ plotted on a Cartesian plane

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 x ↦ x³ : R → R 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 F_p for such prime p that p ≠ 1 (mod 3), 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: x³ − x = 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). Cubic equations were known to the ancient Greek mathematician Diophantus. Hero of Alexandria devised a method for calculating cube roots in the 1st century CE. 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.

See also

• Cabtaxi number
• Cubic equation
• Doubling the cube
• Euler's sum of powers conjecture
• Fifth power (algebra)
• Fourth power
• Kepler's laws of planetary motion#Third law
• Monkey saddle
• Perfect power
• Taxicab number

References

Bibliography

• Hardy, G. H.; Wright, E. M. (1980). An Introduction to the Theory of Numbers (Fifth ed.). Oxford: Oxford University Press. ISBN 978-0-19-853171-5.
• Wheatstone, C. (1854). "On the formation of powers from arithmetical progressions". Proceedings of the Royal Society of London. 7: 145–151. Bibcode:1854RSPS....7..145W. doi:10.1098/rspl.1854.0036.