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 : Geometrically speaking, a positive integer is a perfect cube
if and only if one can arrange 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 . The difference between the cubes of consecutive integers can be expressed as follows: :. or :. There is no minimum perfect cube, since the cube of a negative integer is negative. For example, .
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,
2,
4,
6 and
8 can be the last two digits of a perfect cube (where stands for any odd digit and for any even digit). Some cube numbers are also square numbers; for example, 64 is a square number and a cube number . This happens if and only if the number is a perfect sixth power (in this case 2). 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, • : \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 2 \pmod 3 \quad \text{then} \quad x^3\equiv 8 \pmod 9.
Sums of two cubes Sums of three cubes It is conjectured that every integer (positive or negative) not
congruent to modulo 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. Integers congruent to modulo 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: : 42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3. One solution to x^3 + y^3 + z^3 = n is given in the table below for , and not congruent to or modulo . The selected solution is the one that is primitive (), 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 , and has minimal values for and (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 . For example, for , 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 , the solution is excluded, and this is the solution that is selected.
Fermat's Last Theorem for cubes The equation has no non-trivial (i.e. ) solutions in integers. In fact, it has none in
Eisenstein integers. Both of these statements are also true for the equation .
Sum of first n cubes The sum of the first cubes is the th
triangle number squared: :1^3+2^3+\dots+n^3 = (1+2+\dots+n)^2=\left(\frac{n(n+1)}{2}\right)^2.
Proofs. 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}} (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 = \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} In the more recent mathematical literature, uses the rectangle-counting interpretation of these numbers to form a geometric proof of the identity (see also ); he observes that it may also be proved easily (but uninformatively) by induction, and states that provides "an interesting old Arabic proof". provides a purely visual proof, provide two additional proofs, and 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
odd cubes, :1^3+3^3+\dots+(2y-1)^3 = (xy)^2 but , must satisfy the negative
Pell equation . For example, for and , 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 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 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 for finding the sum of cubes of numbers in arithmetic progression with common difference and initial cube , :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^2n+d^2n^2) A parametric solution to :F(d,a,n) = y^3 is known for the special case of , or consecutive cubes, as found by Pagliani in 1829.
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 (); the sum of the next
two is the next cube (); the sum of the next
three is the next cube (); and so forth.
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. ==In rational numbers==