Cube (algebra)

{{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 {{OEIS|id=A000578}}:

Geometrically speaking, a positive integer {{mvar|m}} is a perfect cube if and only if one can arrange {{mvar|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 {{nowrap|3 × 3 × 3 {{=}} 27}}.

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

:{{math|size=120%|1=n³ − (n − 1)³ = 3(n − 1)n + 1}}.

or

:{{math|size=120%|1=(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, {{nowrap|(−4) × (−4) × (−4) {{=}} −64}}.

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, {{math|o}}2, {{math|e}}4, {{math|o}}6 and {{math|e}}8 can be the last two digits of a perfect cube (where {{math|o}} stands for any odd digit and {{math|e}} for any even digit). Some cube numbers are also square numbers; for example, 64 is a square number {{nowrap|(8 × 8)}} and a cube number {{nowrap|(4 × 4 × 4)}}. This happens if and only if the number is a perfect sixth power (in this case 2{{sup|6}}).

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

{{main|Sum of two cubes}}

{{main|Sums of three cubes}}

It is conjectured that every integer (positive or negative) not congruent to {{math|±4}} modulo {{math|9}} can be written as a sum of three (positive or negative) cubes with infinitely many ways.{{cite arXiv |last=Huisman |first=Sander G. |eprint=1604.07746 |title=Newer sums of three cubes |class=math.NT |date=27 Apr 2016 }} For example, $6\; =\; 2^3+(-1)^3+(-1)^3$. Integers congruent to {{math|±4}} modulo {{math|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:{{cite journal

| last1 = Booker

| first1 = Andrew R.

| last2 = Sutherland

| first2 = Andrew V.

| title = On a question of Mordell

| journal = Proceedings of the National Academy of Sciences

| year = 2021

| volume = 118

| issue = 11

| doi = 10.1073/pnas.2022377118

| pmid = 33692126

| arxiv = 2007.01209

| doi-access = free

| pmc = 7980389

| bibcode = 2021PNAS..11822377B

}}

:$42\; =\; (-80538738812075974)^3\; +\; 80435758145817515^3\; +\; 12602123297335631^3.$

One solution to $x^3\; +\; y^3\; +\; z^3\; =\; n$ is given in the table below for {{math|n ≤ 78}}, and {{math|n}} not congruent to {{math|4}} or {{math|5}} modulo {{math|9}}. The selected solution is the one that is primitive ({{math|1=gcd(x, y, z) = 1}}), is not of the form $c^3+(-c)^3+n^3=n^3$ or $(n+6nc^3)^3+(n-6nc^3)^3+(-6nc^2)^3=2n^3$ (since they are infinite families of solutions), satisfies {{math|0 ≤ {{abs|x}} ≤ {{abs|y}} ≤ {{abs|z|}}}}, and has minimal values for {{math|{{abs|z}}}} and {{math|{{abs|y}}}} (tested in this order).Sequences {{OEIS link|A060465}}, {{OEIS link|A060466}} and {{OEIS link|A060467}} in OEIS[http://cr.yp.to/threecubes/20010729 Threecubes][http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math04/matb0100.htm n=x^3+y^3+z^3]

Only primitive solutions are selected since the non-primitive ones can be trivially deduced from solutions for a smaller value of {{mvar|n}}. For example, for {{math|1=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 {{math|1=n = 48}}, the solution {{math|1=(x, y, z) = (-2, -2, 4)}} is excluded, and this is the solution {{math|1=(x, y, z) = (-23, -26, 31)}} that is selected.

{{sums of three cubes table}}

{{main|Fermat's Last Theorem}}

The equation {{math|1=x³ + y³ = z³}} has no non-trivial (i.e. {{math|xyz ≠ 0}}) solutions in integers. In fact, it has none in Eisenstein integers.Hardy & Wright, Thm. 227

Both of these statements are also true for the equationHardy & Wright, Thm. 232 {{math|1=x³ + y³ = 3z³}}.

The sum of the first {{mvar|n}} cubes is the {{mvar|n}}th triangle number squared:

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

File:Nicomachus_theorem_3D.svg

Proofs.

{{harvs|txt|first=Charles|last=Wheatstone|authorlink=Charles Wheatstone|year=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

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

That identity is related to triangular numbers $T\_n$ in the following way:

:$n^3\; =\backslash sum\; \_\{k=T\_\{n-1\}+1\}^\{T\_\{n\}\}\; (2\; k-1),$

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:

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

we obtain the following derivation:

:$$

\begin{align}

\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{align}

File:Sum of cubes2.png

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

:$1^3+3^3+\backslash dots+(2y-1)^3\; =\; (xy)^2$

but {{mvar|x}}, {{mvar|y}} must satisfy the negative Pell equation {{math|x{{sup|2}} − 2y{{sup|2}} {{=}} −1}}. For example, for {{math|1=y = 5}} and {{math|29}}, then,

:$1^3+3^3+\backslash dots+9^3\; =\; (7\backslash cdot\; 5)^2$

:$1^3+3^3+\backslash dots+57^3\; =\; (41\backslash cdot\; 29)^2$

and so on. Also, every even perfect number, except the lowest, is the sum of the first {{math|power of two}} odd cubes (p = 3, 5, 7, ...):

:$28\; =\; 2^2(2^3-1)\; =\; 1^3+3^3$

:$496\; =\; 2^4(2^5-1)\; =\; 1^3+3^3+5^3+7^3$

:$8128\; =\; 2^6(2^7-1)\; =\; 1^3+3^3+5^3+7^3+9^3+11^3+13^3+15^3$

File:Plato_number.svg

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 {{mvar|F}} for finding the sum of {{mvar|n}}

cubes of numbers in arithmetic progression with common difference {{mvar|d}} and initial cube {{math|a³}},

:$F(d,a,n)\; =\; a^3+(a+d)^3+(a+2d)^3+\backslash cdots+(a+dn-d)^3$

is given by

:$F(d,a,n)\; =\; (n/4)(2a-d+dn)(2a^2-2ad+2adn-d^2n+d^2n^2)$

A parametric solution to

:$F(d,a,n)\; =\; y^3$

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

| last1 = Bennett | first1 = Michael A.

| last2 = Patel | first2 = Vandita

| last3 = Siksek | first3 = Samir

| doi = 10.1112/S0025579316000231

| issue = 1

| journal = Mathematika

| mr = 3610012

| pages = 230–249

| title = Perfect powers that are sums of consecutive cubes

| volume = 63

| year = 2017| arxiv = 1603.08901

}}

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

{{main|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.

Every positive rational number is the sum of three positive rational cubes,Hardy & Wright, Thm. 234 and there are rationals that are not the sum of two rational cubes.Hardy & Wright, Thm. 233

{{more information|cubic function}}

Image:X cubed plot.svg

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 {{math|x ↦ x³ : R → R}} is a surjection (takes all possible values). Only three numbers are equal to their own cubes: {{num|−1}}, {{num|0}}, and {{num|1}}. If {{math|−1 < x < 0}} or {{math|1 < x}}, then {{math|x³ > x}}. If {{math|x < −1}} or {{math|0 < x < 1}}, then {{math|x³ < x}}. All aforementioned properties pertain also to any higher odd power ({{math|x⁵}}, {{math|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, {{math|1=i³ = −i}}.

The derivative of {{math|x³}} equals {{math|3x²}}.

Cubes occasionally have the surjective property in other fields, such as in {{math|F_p}} for such prime {{mvar|p}} that {{math|p ≠ 1 (mod 3)}},The multiplicative group of {{math|F_p}} is cyclic of order {{math|p − 1}}, and if it is not divisible by 3, then cubes define a group automorphism. but not necessarily: see the counterexample with rationals above. Also in {{math|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: {{math|1=x³ − x = x(x − 1)(x + 1)}}.

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).{{cite book|last=Cooke|first=Roger|title=The History of Mathematics|url=https://books.google.com/books?id=CFDaj0WUvM8C&pg=PT63|date=8 November 2012|publisher=John Wiley & Sons|isbn=978-1-118-46029-0|page=63}}{{cite book|last= Nemet-Nejat|first=Karen Rhea|title=Daily Life in Ancient Mesopotamia|url=https://archive.org/details/dailylifeinancie00neme|url-access= registration|year=1998|publisher=Greenwood Publishing Group|isbn=978-0-313-29497-6|page=[https://archive.org/details/dailylifeinancie00neme/page/306 306]}} Cubic equations were known to the ancient Greek mathematician Diophantus.Van der Waerden, Geometry and Algebra of Ancient Civilizations, chapter 4, Zurich 1983 {{ISBN|0-387-12159-5}} Hero of Alexandria devised a method for calculating cube roots in the 1st century CE.{{cite journal|last=Smyly|first=J. Gilbart|title=Heron's Formula for Cube Root|journal=Hermathena|year=1920|volume=19|issue=42|pages=64–67|publisher=Trinity College Dublin|jstor=23037103}} 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.{{cite book|last1=Crossley|first1=John|last2=W.-C. Lun|first2=Anthony|title=The Nine Chapters on the Mathematical Art: Companion and Commentary|url=https://books.google.com/books?id=eiTJHRGTG6YC&pg=PA213|year=1999|publisher=Oxford University Press|isbn=978-0-19-853936-0|pages=176, 213}}

{{Div col}}

• 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

{{Div col end}}

{{Notelist}}

{{Reflist}}

{{refbegin|2}}

• {{cite journal |last1=Benjamin |first1=Arthur T. |last2=Orrison |first2=Michael E. |title=Two Quick Combinatorial Proofs of Σ k = 1 n k 3 = ( \smallmatrix n+1 2 \endsmallmatrix ) 2 |journal=The College Mathematics Journal |date=November 2002 |volume=33 |issue=5 |pages=406 |url=http://www.math.hmc.edu/~orrison/research/papers/two_quick.pdf |doi=10.2307/1559017 |jstor=1559017}}
• {{cite journal |last1=Benjamin |first1=Arthur T. |last2=Quinn |first2=Jennifer J. |last3=Wurtz |first3=Calyssa |title=Summing Cubes by Counting Rectangles |journal=The College Mathematics Journal |date=1 November 2006 |volume=37 |issue=5 |pages=387–389 |doi=10.2307/27646391 |jstor=27646391|url=https://scholarship.claremont.edu/hmc_fac_pub/567 }}
• {{Cite book |last1=Hardy |first1=G. H. |author-link1=G. H. Hardy |last2=Wright |first2=E. M. |author-link2=E. M. Wright |date=1980 |title=An Introduction to the Theory of Numbers |edition=Fifth |publisher=Oxford University Press |location=Oxford |isbn=978-0-19-853171-5 |url-access=registration |url=https://archive.org/details/introductiontoth00hard}}
• {{cite journal |last1=Kanim |first1=Katherine |title=Proof without Words: The Sum of Cubes: An Extension of Archimedes' Sum of Squares |journal=Mathematics Magazine |date=1 October 2004 |volume=77 |issue=4 |pages=298–299 |doi=10.2307/3219288 |jstor=3219288}}
• {{cite book |last1=Nelsen |first1=Roger B. |title=Proofs without words : exercises in visual thinking |date=1993 |publisher=Cambridge University Press |isbn=978-0-88385-700-7}}
• {{cite journal |last1=Stein |first1=Robert G. |title=A Combinatorial Proof That Σ k3 = (Σ k)2 |journal=Mathematics Magazine |date=1 May 1971 |volume=44 |issue=3 |pages=161–162 |doi=10.2307/2688231 |jstor=2688231}}
• {{cite book |last1=Toeplitz |first1=Otto |author-link=Otto Toeplitz |title=The calculus: a genetic approach |date=1963 |publisher=University of Chicago Press |location=Chicago |isbn=978-0-226-80667-9}}
• {{Cite journal |last=Wheatstone |first=C. |author-link=Charles Wheatstone |title=On the formation of powers from arithmetical progressions |journal=Proceedings of the Royal Society of London |volume=7 |pages=145–151 |year=1854 |doi=10.1098/rspl.1854.0036 |bibcode=1854RSPS....7..145W |s2cid=121885197 |doi-access=}}

{{refend}}

{{Figurate numbers}}

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