Cubes 1 to 30: Full Table, Patterns and Quick Methods

What Is A Cube Number?

A cube number (or perfect cube) is a whole number you get by multiplying a whole number by itself three times: $n^3 = n \times n \times n$. The small raised $3$ is an exponent telling you how many copies of $n$ to multiply, which the exponents page explains in general.

The name comes from geometry: $n^3$ unit cubes stack into a solid cube with edge $n$. The live page on cube numbers carries the idea beyond 30.

Three things look alike and are not:

• $3n$ is three copies of $n$ added: $3 \times 4 = 12$.

• $n^2$ is $n$ times itself once: $4^2 = 16$.

• $n^3$ is $n$ times itself twice: $4^3 = 64$.

The cube of a negative number is negative, since an odd count of negative signs leaves one over: $(-3)^3 = -27$.

How Do You Read The Cubes 1 To 30 Table?

Read the table across: find $n$ in the first column, and its cube sits in the last column. For example, the row for $17$ shows $17^3 = 4{,}913$.

Read it backwards to find a cube root. If a problem gives you $13{,}824$, find it in the cube column and read $24$ from the same row, so $\sqrt[3]{13824} = 24$.

The table only lists perfect cubes. A number such as $20{,}000$ is missing because it sits between $27^3 = 19{,}683$ and $28^3 = 21{,}952$, so its cube root is between $27$ and $28$ (about $27.14$).

Cubes Of 1 To 30 Table

Each table lists the number $n$, its square $n^2$ and its cube $n^3$, so you can see how much faster cubes grow than squares.

Cubes From 1 To 10

Cubes From 11 To 20

Cubes From 21 To 30

The squares column is the same list you will find on square 1 to 30, and multiplying each square by $n$ once more gives the cube beside it.

What Are The Cubes Of Even And Odd Numbers From 1 To 30?

The cubes of the fifteen even numbers are all even, and the cubes of the fifteen odd numbers are all odd. Cubing keeps the factor $2$ if it is there and cannot create one if it is not.

Every even cube is in fact a multiple of $8$, because $(2k)^3 = 8k^3$. So $22^3 = 8 \times 11^3 = 8 \times 1{,}331 = 10{,}648$.

Where Do Cubes 1 To 30 Show Up, And Why Do They Matter?

Cubes 1 to 30 show up whenever you measure the space inside a box with equal sides. A cube-shaped box $17$ cm on each side holds $17^3 = 4{,}913$ cubic centimetres, which is just under $5$ litres.

The litre itself is a cube: a box $10$ cm on each side holds $10^3 = 1{,}000\ \text{cm}^3$, exactly one litre. A $3 \times 3 \times 3$ puzzle cube is built from $3^3 = 27$ small cubes, and doubling every side of any box multiplies its volume by $2^3 = 8$, which is why a box twice as wide holds eight times as much.

What Patterns Hide In Cubes 1 To 30?

Five patterns run through the table, and each one lets you rebuild or check a cube instead of recalling it cold.

Last Digits Repeat Every Ten

The last digit of $n^3$ depends only on the last digit of $n$:

Seven digits map to themselves; only the pairs $2 \leftrightarrow 8$ and $3 \leftrightarrow 7$ swap. So $12^3$ ends in $8$ and $28^3$ ends in $2$, and you can check either one in a second.

Digital Roots Are Always 1, 8 Or 9

Add the digits of any cube, and keep adding until one digit is left. The result is always $1$, $8$ or $9$, in the repeating order $1, 8, 9$ as $n$ goes $1, 2, 3, 4, \ldots$

For $24^3 = 13{,}824$: $1 + 3 + 8 + 2 + 4 = 18$, then $1 + 8 = 9$. Since $24$ is a multiple of $3$, the digital root $9$ is the expected one, which makes this a fast check on a calculation.

The Gaps Between Cubes Grow Steadily

Consecutive cubes differ by $3n^2 + 3n + 1$:

$$(n + 1)^3 - n^3 = 3n^2 + 3n + 1$$

The gaps run $7, 19, 37, 61, 91, \ldots$ For the last step in the table, $30^3 - 29^3 = 27{,}000 - 24{,}389 = 2{,}611$, and $3(29)^2 + 3(29) + 1 = 2{,}611$ as well.

Every Cube Is A Run Of Odd Numbers

Write the odd numbers in order and group them as $1$, then the next two, then the next three. Each group adds up to a cube:

$$1 = 1^3$$

$$3 + 5 = 8 = 2^3$$

$$7 + 9 + 11 = 27 = 3^3$$

$$13 + 15 + 17 + 19 = 64 = 4^3$$

In general $n^3$ is the sum of $n$ consecutive odd numbers starting at $n^2 - n + 1$, so $30^3$ is $871 + 873 + \cdots + 929$, thirty odd numbers with total $27{,}000$.

Some Cubes Are Also Squares

When $n$ is itself a perfect square, $n^3$ is a perfect square too. Inside 1 to 30 that gives five numbers that are both: $1$, $64 = 8^2$, $729 = 27^2$, $4{,}096 = 64^2$ and $15{,}625 = 125^2$.

What Is The Sum Of Cubes From 1 To 30?

The sum of cubes from 1 to 30 is $216{,}225$. You do not need to add thirty numbers, because the sum of the first $n$ cubes equals the square of the sum of the first $n$ whole numbers:

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

$$1^3 + 2^3 + \cdots + 30^3 = \left(\frac{30 \times 31}{2}\right)^2 = 465^2 = 216{,}225$$

The inner part, $\frac{n(n+1)}{2}$, is the triangular number from sum of natural numbers. The proof and more cases are on sum of cubes of n natural numbers.

How Do You Remember Cubes From 1 To 30?

You remember cubes from 1 to 30 best by knowing ten of them well and rebuilding the other twenty from patterns. A chanted list tends to fade, while a method you understand can always be rebuilt.

1. Anchor on 1 to 10. These ten ($1, 8, 27, 64, 125, 216, 343, 512, 729, 1{,}000$) are the ones worth knowing on sight, because every later cube is built from them.

2. Use the decade anchors. $10^3 = 1{,}000$, $20^3 = 8{,}000$, $30^3 = 27{,}000$: the cube of a multiple of ten is the cube of its first digit with three zeros.

3. Check the last digit. The last-digit table above confirms every answer, and it is right for all thirty.

4. Step from the nearest anchor. Numbers next to $10$, $20$ or $30$ come straight from the identity in the next section.

5. Check with the digital root. If your answer's digital root is not $1$, $8$ or $9$, something slipped.

How Do You Calculate A Cube Quickly In Your Head?

You calculate a cube in your head by splitting the number into a round part and a small part, then using the cube identity. The identity is the cube of a binomial:

$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$

$$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$$

For $21^3$, take $a = 20$, $b = 1$:

$$21^3 = 8{,}000 + 3(400)(1) + 3(20)(1) + 1$$

$$= 8{,}000 + 1{,}200 + 60 + 1$$

$$= 9{,}261$$

For $29^3$, take $a = 30$, $b = 1$ with the minus version:

$$29^3 = 27{,}000 - 2{,}700 + 90 - 1$$

$$= 24{,}389$$

For a number in the middle of a decade, squaring first is often faster: $15^3 = 225 \times 15 = 3{,}375$, using $225 \times 15 = 2{,}250 + 1{,}125$. The mental math tricks page collects more splits like this one.

How Do You Find A Cube Root From The Table?

You find the cube root of any perfect cube up to $27{,}000$ from two clues: its last digit and its size. This is the "cube root 1 to 30 trick" in step-by-step form.

1. Last digit of the root: Use the last-digit table backwards: a cube ending in $7$ has a root ending in $3$, ending in $2$ means a root ending in $8$, and so on.

2. Tens digit of the root: Drop the last three digits of the cube and see which decade cube the rest sits above: $1$ to $7$ means the root is in the teens, $8$ to $26$ means the twenties.

3. Put them together: For $12{,}167$: last digit $7$ gives $3$; dropping $167$ leaves $12$, which is between $8$ and $27$, so the tens digit is $2$. The root is $23$.

Check: $23^3 = 12{,}167$. The full list of roots is on cube root 1 to 100.

Examples Of Cubes 1 To 30

Example 1

Find the volume of a cube-shaped box with edge $14$ cm.

$$V = 14^3$$

$$= 196 \times 14$$

$$= 2{,}744\ \text{cm}^3$$

Final answer: $2{,}744\ \text{cm}^3$

Example 2

A student finds $21^3$ by writing $(20 + 1)^3 = 20^3 + 1^3 = 8{,}001$. Is that right?

Check the size first. $21^3$ must be more than $20^3 = 8{,}000$ by a lot more than $1$, because one extra unit on every edge adds three whole $20 \times 20$ slabs of volume.

The break is splitting the cube over the plus sign: $(a + b)^3$ is not $a^3 + b^3$. The two middle terms $3a^2b = 1{,}200$ and $3ab^2 = 60$ went missing.

Final answer: $21^3 = 8{,}000 + 1{,}200 + 60 + 1 = 9{,}261$.

Example 3

Which number has a cube of $13{,}824$?

Last digit $4$ gives a root ending in $4$. Dropping $824$ leaves $13$, between $8$ and $27$, so the tens digit is $2$.

$$24^3 = 576 \times 24 = 13{,}824$$

Final answer: $24$

Example 4

Find $1^3 + 2^3 + \cdots + 10^3$, then the sum of cubes from 11 to 30.

$$1^3 + \cdots + 10^3 = 55^2 = 3{,}025$$

$$11^3 + \cdots + 30^3 = 216{,}225 - 3{,}025 = 213{,}200$$

Final answer: $3{,}025$ and $213{,}200$

Readers who add the ten cubes one by one usually land within a few units of $3{,}025$ and then cannot tell which addition slipped; the square-of-55 route removes that risk entirely.

Example 5

Show that $1{,}729$ is a sum of two cubes in two different ways, using only the table.

$$1^3 + 12^3 = 1 + 1{,}728 = 1{,}729$$

$$9^3 + 10^3 = 729 + 1{,}000 = 1{,}729$$

Final answer: both pairs, $(1, 12)$ and $(9, 10)$, give $1{,}729$.

Example 6

How many dice with $3$ cm edges fill a cube-shaped box with $30$ cm edges?

Each edge holds $\frac{30}{3} = 10$ dice, so the box holds

$$10^3 = 1{,}000 \text{ dice}$$

Check by volume: $\frac{30^3}{3^3} = \frac{27{,}000}{27} = 1{,}000$.

Final answer: $1{,}000$ dice

What Mistakes Do Students Make With Cubes 1 To 30?

Mistake 1: Multiplying By 3 Instead Of Cubing

Where it slips in: The first week of exponents, when the raised $3$ looks like an ordinary factor.

Don't do this: Write $4^3 = 12$.

The correct way: $4^3 = 4 \times 4 \times 4 = 64$. Students who say "four times four times four" aloud for the first few problems stop confusing the exponent with a multiplier.

Mistake 2: Dropping A Zero On The Decade Cubes

Where it slips in: Cubing $10$, $20$ or $30$ quickly, when the zeros get counted as two instead of three.

Don't do this: Write $30^3 = 2{,}700$.

The correct way: Each factor of $10$ brings one zero, and there are three factors: $30^3 = 27 \times 1{,}000 = 27{,}000$. The rusher who writes $2{,}700$ has actually computed $3 \times 30^2$, the same $2{,}700$ that appears in the expansion of $29^3$.

Mistake 3: Swapping The 2 And 8 (Or 3 And 7) Endings

Where it slips in: Checking a cube's last digit from memory, especially for $12$, $18$, $22$ and $28$.

Don't do this: Expect $12^3$ to end in $2$, then "correct" a right answer of $1{,}728$.

The correct way: Only two pairs swap: $2 \leftrightarrow 8$ and $3 \leftrightarrow 7$. The second-guesser who doubts a correct $1{,}728$ is the reader this pair trips most, so check with $2 \times 2 \times 2 = 8$ instead of trusting a hunch.

Where Can You Get Help With Cubes And Mental Maths?

Cubes become automatic once someone has you rebuild them from patterns a few times with feedback. Mental maths for kids practises exactly this kind of quick, pattern-based arithmetic, and a middle school math tutor can connect cubes to volume and exponents.

For learners moving into identities like $(a + b)^3$, an algebra tutor helps, and math classes online give a steady weekly routine.

Conclusion

• Cubes 1 to 30 are the thirty perfect cubes from $1$ to $27{,}000$, with $n^3 = n \times n \times n$.

• Even numbers have even cubes and odd numbers have odd cubes; the last digit and the digital root check any value.

• The sum of all thirty is $465^2 = 216{,}225$, from the sum-of-cubes formula.

• The $(a \pm b)^3$ identity rebuilds any cube near $10$, $20$ or $30$ in seconds.

• The usual errors are writing $3n$ for $n^3$, dropping a zero on decade cubes, and swapping the 2 and 8 endings.

Practice These Before Moving On

1. Find $19^3$ using $(20 - 1)^3$.

2. Which number has a cube of $17{,}576$?

3. Find $11^3 + 12^3 + 13^3$.

4. A cube has volume $3{,}375\ \text{cm}^3$. Find its edge.

5. Find the difference $26^3 - 25^3$ without working out either cube.

Answers:

1. $8{,}000 - 1{,}200 + 60 - 1 = 6{,}859$

2. $26$ (last digit $6$ gives $6$; $17$ lies between $8$ and $27$)

3. $1{,}331 + 1{,}728 + 2{,}197 = 5{,}256$

4. $15$ cm

5. $3(25)^2 + 3(25) + 1 = 1{,}951$

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Read More

• Perfect cube: definition and test

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• Cube root of 343

• The a cube plus b cube formula

• Perfect squares and their patterns