Why Is Every Cube 0, 1, or 8 Modulo 9?

To prove that an integer cube has the form $9k$, $9k+1$, or $9k+8$, split the integers into their three possible remainders upon division by 3. Cubing each of those forms shows that the result differs from one of these three numbers by a multiple of 9.

Here $k$ must be an integer, and the claim applies to negative integers as well as positive ones.

Proof by cases modulo 3

By the division algorithm, every integer $n$ can be written in exactly one of these ways for some integer $q$:

$$n=3q, \qquad n=3q+1, \qquad \text{or} \qquad n=3q+2.$$

These are the only cases because the possible remainders after division by 3 are 0, 1, and 2.

Case 1: $n=3q$

Cubing gives

$$n^3=(3q)^3=27q^3=9(3q^3).$$

Since $3q^3$ is an integer, this has the form $9k$.

Case 2: $n=3q+1$

Expand the cube:

$$\begin{aligned} n^3&=(3q+1)^3\\ &=27q^3+27q^2+9q+1\\ &=9(3q^3+3q^2+q)+1. \end{aligned}$$

The quantity in parentheses is an integer, so this has the form $9k+1$.

Case 3: $n=3q+2$

Similarly,

$$\begin{aligned} n^3&=(3q+2)^3\\ &=27q^3+54q^2+36q+8\\ &=9(3q^3+6q^2+4q)+8. \end{aligned}$$

Again, the quantity in parentheses is an integer. This has the form $9k+8$.

In all three cases, $k$ is an integer. Therefore every integer cube is congruent to $0$, $1$, or $8$ modulo 9:

$$n^3\equiv 0,\ 1,\ \text{or }8\pmod 9.$$

Worked examples

Take $n=6$. It is divisible by 3, so the first case applies:

$$6^3=216=9\cdot24.$$

For $n=4$, write $4=3(1)+1$. The second case predicts a remainder of 1, and indeed

$$4^3=64=9\cdot7+1.$$

For $n=5$, write $5=3(1)+2$. The third case predicts a remainder of 8:

$$5^3=125=9\cdot13+8.$$

The same reasoning handles negative inputs. For example, $-4=3(-2)+2$, so its cube must have remainder 8 modulo 9. In fact,

$$(-4)^3=-64=9(-8)+8.$$

Why checking remainders modulo 3 is enough

The proof works because the three possible remainders modulo 3 cover every integer, and cubing each representative gives the corresponding remainder modulo 9:

Remainder of $n$ modulo 3	Remainder of $n^3$ modulo 9
0	0
1	1
2	8

For the last row, $2^3=8$. More generally, replacing $n$ by another integer with the same remainder modulo 3 does not change which of the three cube forms applies; the expansions above show the difference is a multiple of 9.

Common mistakes

• Checking only positive numbers: The division algorithm and the proof hold for every integer, including negative values.
• Using $k$ as if it had to be positive: The statement only requires $k$ to be an integer. For instance, $(-4)^3=9(-8)+8$ uses negative $k$.
• Writing the last form as $9k-1$ without explaining it: This is equivalent to $9k+8$, since $9k+8=9(k+1)-1$. Both notations describe the same remainder class modulo 9.
• Claiming every number of these forms is a cube: The proof shows that cubes must have one of these remainders. It does not say that every integer congruent to 0, 1, or 8 modulo 9 is itself a cube.