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In integers

See also: Cube-free integer

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):

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 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, if x ≡ 0 ( mod 3 ) then x 3 ≡ 0 ( mod 9 )  (actually 0 ( mod 27 ) ) ; {\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, if x ≡ 1 ( mod 3 ) then x 3 ≡ 1 ( mod 9 ) ; {\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, if x ≡ 2 ( mod 3 ) then x 3 ≡ 8 ( mod 9 ) . {\displaystyle {\text{if}}\quad x\equiv 2{\pmod {3}}\quad {\text{then}}\quad x^{3}\equiv 8{\pmod {9}}.}

Sums of two cubes

Main article: Sum of two cubes

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.² For example, 6 = 2 3 + ( − 1 ) 3 + ( − 1 ) 3 {\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:³

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

One solution to x 3 + y 3 + z 3 = n {\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 c 3 + ( − c ) 3 + n 3 = n 3 {\displaystyle c^{3}+(-c)^{3}+n^{3}=n^{3}} or ( n + 6 n c 3 ) 3 + ( n − 6 n c 3 ) 3 + ( − 6 n c 2 ) 3 = 2 n 3 {\displaystyle (n+6nc^{3})^{3}+(n-6nc^{3})^{3}+(-6nc^{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).⁴⁵⁶

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 {\displaystyle 2^{3}+2^{3}+2^{3}=24} results from the solution 1 3 + 1 3 + 1 3 = 3 {\displaystyle 1^{3}+1^{3}+1^{3}=3} by multiplying everything by 8 = 2 3 . {\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.

Primitive solutions for n from 1 to 78
n	x	y	z		n	x	y	z
1	9	10	−12		39	117367	134476	−159380
2	1214928	3480205	−3528875		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

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.⁷

Both of these statements are also true for the equation⁸ x3 + y3 = 3z3.

Sum of first n cubes

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

1 3 + 2 3 + ⋯ + n 3 = ( 1 + 2 + ⋯ + n ) 2 = ( n ( n + 1 ) 2 ) 2 . {\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

n 3 = ( n 2 − n + 1 ) + ( n 2 − n + 1 + 2 ) + ( n 2 − n + 1 + 4 ) + ⋯ + ( n 2 + n − 1 ) ⏟ n  consecutive odd numbers . {\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 T n {\displaystyle T_{n}} in the following way:

n 3 = ∑ k = T n − 1 + 1 T n ( 2 k − 1 ) , {\displaystyle n^{3}=\sum _{k=T_{n-1}+1}^{T_{n}}(2k-1),}

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

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

we obtain the following derivation:

∑ k = 1 n k 3 = 1 + 8 + 27 + 64 + ⋯ + n 3 = 1 ⏟ 1 3 + 3 + 5 ⏟ 2 3 + 7 + 9 + 11 ⏟ 3 3 + 13 + 15 + 17 + 19 ⏟ 4 3 + ⋯ + ( n 2 − n + 1 ) + ⋯ + ( n 2 + n − 1 ) ⏟ n 3 = 1 ⏟ 1 2 + 3 ⏟ 2 2 + 5 ⏟ 3 2 + ⋯ + ( n 2 + n − 1 ) ⏟ ( n 2 + n 2 ) 2 = ( 1 + 2 + ⋯ + n ) 2 = ( ∑ k = 1 n k ) 2 . {\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,

1 3 + 2 3 + 3 3 + 4 3 + 5 3 = 15 2 {\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,

1 3 + 3 3 + ⋯ + ( 2 y − 1 ) 3 = ( x y ) 2 {\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,

1 3 + 3 3 + ⋯ + 9 3 = ( 7 ⋅ 5 ) 2 {\displaystyle 1^{3}+3^{3}+\dots +9^{3}=(7\cdot 5)^{2}} 1 3 + 3 3 + ⋯ + 57 3 = ( 41 ⋅ 29 ) 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, ...):

28 = 2 2 ( 2 3 − 1 ) = 1 3 + 3 3 {\displaystyle 28=2^{2}(2^{3}-1)=1^{3}+3^{3}} 496 = 2 4 ( 2 5 − 1 ) = 1 3 + 3 3 + 5 3 + 7 3 {\displaystyle 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 {\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:

3 3 + 4 3 + 5 3 = 6 3 {\displaystyle 3^{3}+4^{3}+5^{3}=6^{3}} 11 3 + 12 3 + 13 3 + 14 3 = 20 3 {\displaystyle 11^{3}+12^{3}+13^{3}+14^{3}=20^{3}} 31 3 + 33 3 + 35 3 + 37 3 + 39 3 + 41 3 = 66 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,

F ( d , a , n ) = a 3 + ( a + d ) 3 + ( a + 2 d ) 3 + ⋯ + ( a + d n − d ) 3 {\displaystyle 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 ) ( 2 a − d + d n ) ( 2 a 2 − 2 a d + 2 a d n − d 2 n + d 2 n 2 ) {\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

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

is known for the special case of d = 1, or consecutive cubes, as found by Pagliani in 1829.⁹

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

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.

In rational numbers

Every positive rational number is the sum of three positive rational cubes,¹⁰ and there are rationals that are not the sum of two rational cubes.¹¹

In real numbers, other fields, and rings

Further information: cubic function

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 Fp for such prime p that p ≠ 1 (mod 3),¹² 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 their 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).¹³¹⁴ Cubic equations were known to the ancient Greek mathematician Diophantus.¹⁵ Hero of Alexandria devised a method for calculating cube roots in the 1st century CE.¹⁶ 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.¹⁷

See also

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

Notes

Sources

• Benjamin, Arthur T.; Orrison, Michael E. (November 2002). "Two Quick Combinatorial Proofs of Σ k = 1 n k 3 = ( \smallmatrix n+1 2 \endsmallmatrix ) 2" (PDF). The College Mathematics Journal. 33 (5): 406. doi:10.2307/1559017. JSTOR 1559017.
• Benjamin, Arthur T.; Quinn, Jennifer J.; Wurtz, Calyssa (1 November 2006). "Summing Cubes by Counting Rectangles". The College Mathematics Journal. 37 (5): 387–389. doi:10.2307/27646391. JSTOR 27646391.
• 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.
• Kanim, Katherine (1 October 2004). "Proof without Words: The Sum of Cubes: An Extension of Archimedes' Sum of Squares". Mathematics Magazine. 77 (4): 298–299. doi:10.2307/3219288. JSTOR 3219288.
• Nelsen, Roger B. (1993). Proofs without words : exercises in visual thinking. Cambridge University Press. ISBN 978-0-88385-700-7.
• Stein, Robert G. (1 May 1971). "A Combinatorial Proof That Σ k3 = (Σ k)2". Mathematics Magazine. 44 (3): 161–162. doi:10.2307/2688231. JSTOR 2688231.
• Toeplitz, Otto (1963). The calculus: a genetic approach. Chicago: University of Chicago Press. ISBN 978-0-226-80667-9.
• 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. S2CID 121885197.

References

1. The Unicode superscript character ³ is also available for typesetting: n³. /wiki/Unicode_superscript ↩

2. Huisman, Sander G. (27 Apr 2016). "Newer sums of three cubes". arXiv:1604.07746 [math.NT]. /wiki/ArXiv_(identifier) ↩

3. Booker, Andrew R.; Sutherland, Andrew V. (2021). "On a question of Mordell". Proceedings of the National Academy of Sciences. 118 (11). arXiv:2007.01209. Bibcode:2021PNAS..11822377B. doi:10.1073/pnas.2022377118. PMC 7980389. PMID 33692126. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7980389 ↩

4. Sequences A060465, A060466 and A060467 in OEIS //oeis.org/A060465 ↩

5. Threecubes http://cr.yp.to/threecubes/20010729 ↩

6. n=x^3+y^3+z^3 http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math04/matb0100.htm ↩

7. Hardy & Wright, Thm. 227 ↩

8. Hardy & Wright, Thm. 232 ↩

9. Bennett, Michael A.; Patel, Vandita; Siksek, Samir (2017), "Perfect powers that are sums of consecutive cubes", Mathematika, 63 (1): 230–249, arXiv:1603.08901, doi:10.1112/S0025579316000231, MR 3610012 /wiki/ArXiv_(identifier) ↩

10. Hardy & Wright, Thm. 234 ↩

11. Hardy & Wright, Thm. 233 ↩

12. 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. /wiki/Multiplicative_group ↩

13. Cooke, Roger (8 November 2012). The History of Mathematics. John Wiley & Sons. p. 63. ISBN 978-1-118-46029-0. 978-1-118-46029-0 ↩

14. Nemet-Nejat, Karen Rhea (1998). Daily Life in Ancient Mesopotamia. Greenwood Publishing Group. p. 306. ISBN 978-0-313-29497-6. 978-0-313-29497-6 ↩

15. Van der Waerden, Geometry and Algebra of Ancient Civilizations, chapter 4, Zurich 1983 ISBN 0-387-12159-5 /wiki/ISBN_(identifier) ↩

16. Smyly, J. Gilbart (1920). "Heron's Formula for Cube Root". Hermathena. 19 (42). Trinity College Dublin: 64–67. JSTOR 23037103. /wiki/JSTOR_(identifier) ↩

17. 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. 978-0-19-853936-0 ↩