Cubes and Cube Roots: Properties, Methods, Examples

What Are Cubes And Cube Roots?

A cube-shaped water tank that holds exactly 1,000 litres has edges exactly 1 metre long. The volume came first and the edge length came out of it, and that backwards step is a cube root.

Cubes and cube roots are a pair of inverse operations. The cube of a number $a$ is $a^3 = a \times a \times a$, and the cube root of a number $b$ is the number whose cube is $b$, written $\sqrt[3]{b}$.

$$4^3 = 4 \times 4 \times 4 = 64$$

$$\sqrt[3]{64} = 4$$

The small 3 tucked into the root sign is the index. It separates a cube root from the squares and square roots you meet first, where the index 2 is usually left unwritten.

The name is literal. A cube with edge $a$ has volume $a^3$, so "cubing" a number means building a cube from it, and taking the cube root means reading the edge off a known volume.

In exponents language, the cube root is the power one-third. That is why $\sqrt[3]{b} = b^{1/3}$, and why a calculator's $x^{1/3}$ key gives the same answer as its cube root key.

Why Do Cubes And Cube Roots Exist?

Cube roots exist because people needed the edge you need from the space you have. Whenever a volume is fixed by a requirement (a tank, a crate, a block of metal) and the shape is a cube, the only unknown is the edge, and only a cube root can recover it.

• Volume to length: a container must hold 8,000 cubic centimetres, so its edge is $\sqrt[3]{8000} = 20$ cm.

• Scaling: double every edge of a box and its volume grows by $2^3 = 8$, not by 2. Cubes measure how fast volume runs away from length.

• A problem that stumped the Greeks: the Delian problem asked for a cube with twice the volume of a given cubical altar. The new edge must be $\sqrt[3]{2}$ times the old one, and in 1837 Pierre Wantzel proved that length cannot be built with straightedge and compass alone.

That last point shows where the idea leads. Cube roots sit underneath the cubic equations of later algebra, and the cube root of 1 opens the door to complex numbers.

What Are Perfect Cubes?

A perfect cube is a whole number that is the cube of an integer, such as 1, 8, 27 and 64. These are the cube numbers, and their cube roots are exact integers.

Read the table backwards to get cube roots: $\sqrt[3]{4913} = 17$ because $17^3 = 4913$. Every value above was recomputed by script before publishing.

Numbers between two perfect cubes have cube roots between two integers. Since $27 < 30 < 64$, the cube root of 30 lies between 3 and 4, and it is not a whole number. The full cube root 1 to 100 table lists those in-between values to several decimal places.

What Are The Properties Of Cubes?

Cubes follow patterns you can check in seconds, and each one doubles as a quick way to reject a wrong answer.

• Odd stays odd, even stays even: $7^3 = 343$ is odd and $8^3 = 512$ is even.

• Negative cubes are negative: $(-3)^3 = -27$, because three negative signs multiply to a negative.

• Units digits: a number ending in 0, 1, 4, 5, 6 or 9 has a cube ending in the same digit. The pairs 2 and 8, and 3 and 7, swap: $2^3 = 8$, $8^3 = 512$, $3^3 = 27$, $7^3 = 343$.

• Prime factors come in threes: in a perfect cube, every prime in the factorisation has an exponent that is a multiple of 3, as in $1728 = 2^6 \times 3^3$.

• Trailing zeros come in threes: $10^3 = 1000$ and $30^3 = 27000$. A number ending in exactly two zeros can never be a perfect cube.

• Sums of consecutive odd numbers: $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ and $4^3 = 13 + 15 + 17 + 19$.

• Sum of the first $n$ cubes: $1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$. For $n = 4$, $1 + 8 + 27 + 64 = 100 = 10^2$. The sum of cubes of n natural numbers article proves it.

• Digit sums: the repeated digit sum (digital root) of any cube is 1, 8 or 9. If 5,830 shows up as a "cube", its digit sum is 7, so it cannot be one.

What Are The Properties Of Cube Roots?

Cube roots inherit their rules from exponents, because $\sqrt[3]{b} = b^{1/3}$.

• Every real number has exactly one real cube root. Zero's is 0, positive numbers have positive cube roots, and negative numbers have negative cube roots.

• Product rule: $\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$, so $\sqrt[3]{8000} = \sqrt[3]{8} \times \sqrt[3]{1000} = 2 \times 10 = 20$.

• Quotient rule: $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$, so $\sqrt[3]{\frac{27}{125}} = \frac{3}{5}$.

• Undoing: $\sqrt[3]{a^3} = a$ for every real $a$, including negatives. Square roots do not behave this way, since $\sqrt{(-3)^2} = 3$.

• No rule for sums: $\sqrt[3]{a + b}$ is not $\sqrt[3]{a} + \sqrt[3]{b}$.

Beyond the real numbers, every nonzero number also has two complex cube roots. The three cube roots of unity are the standard place to meet them.

How Do You Find The Cube Root Of A Number?

There are four practical routes, and the right one depends on whether the number is a perfect cube and how large it is.

Method 1: Prime Factorisation

Prime factorisation works best for perfect cubes. Break the number into prime numbers, group identical primes in threes, and take one prime from each group.

Find $\sqrt[3]{13824}$:

$$13824 = 2^9 \times 3^3$$

$$= (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)$$

$$\sqrt[3]{13824} = 2 \times 2 \times 2 \times 3 = 24$$

Check: $24^3 = 13824$. If a prime is left over without a full group of three, the number is not a perfect cube.

Method 2: Grouping Digits And The Units Digit

For a perfect cube up to six digits, you can read the cube root almost without arithmetic. Take $91125$:

1. Split the digits into groups of three from the right: $91 \mid 125$.

2. The last group ends in 5, and only cubes of numbers ending in 5 end in 5. So the units digit of the root is 5.

3. Look at the first group, 91. The largest cube not above 91 is $4^3 = 64$, so the tens digit is 4.

4. The cube root is 45. Check: $45^3 = 91125$.

This shortcut assumes the number is already a perfect cube. On a non-cube it produces a confident wrong answer, which is why the final check matters.

How Do You Find A Cube Root Without A Calculator?

For a number that is not a perfect cube, estimate and then refine. The cube root of 30 sits between 3 and 4, since $3^3 = 27$ and $4^3 = 64$.

The tempting start is 3.5, halfway between. A better start is 3, because 30 is only 3 above 27 and a long way below 64, so the answer hugs the lower end. Newton's method then improves any guess $x$:

$$x_{\text{new}} = \frac{1}{3}\left(2x + \frac{30}{x^2}\right)$$

$$x_1 = \frac{1}{3}\left(6 + \frac{30}{9}\right) = 3.1111$$

$$x_2 = \frac{1}{3}\left(6.2222 + \frac{30}{9.6790}\right) = 3.1072$$

Two rounds give $\sqrt[3]{30} \approx 3.1072$, correct to four decimal places, and each round roughly doubles the number of correct digits. (Halley's method, which some textbooks show, converges even faster but needs more arithmetic per step.)

Method 3: Calculator Check

On a scientific calculator, use the $\sqrt[3]{\phantom{x}}$ key, or raise the number to the power $\frac{1}{3}$ with brackets: $30^{(1/3)} = 3.10723\ldots$ Without the brackets, many calculators compute $30^1 \div 3 = 10$, a classic keying error.

Can A Cube Root Be Negative?

Yes. The cube root of a negative number is negative and perfectly real, because a negative number multiplied by itself three times stays negative.

$$(-5) \times (-5) \times (-5) = -125$$

$$\sqrt[3]{-125} = -5$$

So a cube root is not always positive. This is the sharpest difference from square roots: $\sqrt{-125}$ has no real value, because any real number squared is zero or positive.

The graph makes the reason visible. The cube root curve runs through every $x$-value, left and right of zero, while the square root curve stops at zero.

What Is The Difference Between A Cube And A Cube Root?

A cube multiplies a number by itself three times and makes it bigger (for numbers above 1); a cube root asks which number was cubed and makes it smaller. They undo each other.

The square-and-square-root pair works the same way, one power lower. The big change at power three is the sign: cubes keep the sign of the number, squares never return a negative.

Is The Cube Root Of 2 Irrational?

Yes. The cube root of 2, about 1.2599, is an irrational number, and so is the cube root of every positive integer that is not a perfect cube.

The proof mirrors the classic one for $\sqrt{2}$:

1. Suppose $\sqrt[3]{2} = \frac{p}{q}$ in lowest terms.

2. Cubing gives $p^3 = 2q^3$, so $p^3$ is even, which forces $p$ to be even. Write $p = 2k$.

3. Then $8k^3 = 2q^3$, so $q^3 = 4k^3$ and $q$ is even too.

4. Both $p$ and $q$ are even, which contradicts "lowest terms". So $\sqrt[3]{2}$ cannot be a fraction.

This is the same $\sqrt[3]{2}$ the Delian problem asked for. It is a perfectly good number on the number line; it just has no exact fraction or finite decimal form.

Where Are Cubes And Cube Roots Used?

Cubes and cube roots appear wherever three dimensions do, because volume is a length cubed.

• Volume and capacity: the volume formula for a cube is $V = a^3$, and $1000,\text{cm}^3$ is exactly one litre, a $10,\text{cm}$ cube.

• Scaling models: a model built at one-tenth scale has one-thousandth of the original volume, since $\left(\frac{1}{10}\right)^3 = \frac{1}{1000}$.

• Finding a size from a volume: a sphere's radius from its volume is $r = \sqrt[3]{\frac{3V}{4\pi}}$, the same edge-from-volume step.

• Number theory: 1729 is the smallest number that is a sum of two positive cubes in two different ways, $1^3 + 12^3 = 9^3 + 10^3$, which ties straight into the a cube plus b cube formula.

You will meet cube roots again inside cubic equations, in Kepler's laws of planetary motion, and as the power one-third in fractional exponents.

Examples Of Cubes And Cube Roots

Example 1

Find $7^3$ and $(-4)^3$.

$$7^3 = 7 \times 7 \times 7 = 343$$

$$(-4)^3 = (-4) \times (-4) \times (-4) = -64$$

Example 2

Find $\sqrt[3]{8 + 27}$.

Wrong path: split the root across the plus sign.

$$\sqrt[3]{8} + \sqrt[3]{27} = 2 + 3 = 5$$

Check it: $5^3 = 125$, not 35. The answer is far too big, so the split broke something.

Correct path: add first, then take the root.

$$\sqrt[3]{8 + 27} = \sqrt[3]{35} \approx 3.2711$$

Cube roots split across multiplication, never across addition.

Example 3

Is 1188 a perfect cube? If not, find the smallest number to multiply it by to make one.

$$1188 = 2^2 \times 3^3 \times 11$$

The 3s form a full triple, but 2 appears twice and 11 once. To complete the triples, multiply by one more 2 and two more 11s:

$$2 \times 11^2 = 242$$

$$1188 \times 242 = 287496 = 2^3 \times 3^3 \times 11^3 = 66^3$$

So 1188 is not a perfect cube, and 242 is the smallest multiplier.

Example 4

Find $\sqrt[3]{-\frac{3375}{512}}$.

$$3375 = 15^3, \quad 512 = 8^3$$

$$\sqrt[3]{-\frac{3375}{512}} = -\frac{15}{8}$$

The negative sign survives the cube root, and the quotient rule handles the fraction.

Example 5

Find $\sqrt[3]{13824}$ by grouping digits.

Groups from the right: $13 \mid 824$.

The last group ends in 4, so the root ends in 4.

The largest cube not above 13 is $2^3 = 8$, so the tens digit is 2.

The cube root is 24, and $24^3 = 13824$ confirms it. The step students skip here is that final cube; it is the only thing that proves 13824 was a perfect cube to begin with.

Example 6

A cube has edge 10 cm. Find the edge of a second cube with exactly twice its volume.

$$V_1 = 10^3 = 1000,\text{cm}^3$$

$$V_2 = 2000,\text{cm}^3$$

$$a_2 = \sqrt[3]{2000} = 10\sqrt[3]{2} \approx 12.60,\text{cm}$$

Doubling the volume stretches each edge by only about 26%, not 100%. This is the Delian problem with numbers attached.

Where Do Students Go Wrong With Cubes And Cube Roots?

Mistake 1: Multiplying By 3 Instead Of Cubing

Where it slips in: Quick mental work, especially with small numbers.

Don't do this: Writing $4^3 = 12$ or $2^3 = 6$.

The correct way: The exponent counts factors: $4^3 = 4 \times 4 \times 4 = 64$. The rusher's first instinct is to read the little 3 as "times 3", and the habit that fixes it is saying the factors aloud, "four, times four, times four", before writing anything.

Mistake 2: Saying A Negative Number Has No Cube Root

Where it slips in: Right after a unit on square roots.

Don't do this: Writing "no real answer" for $\sqrt[3]{-27}$.

The correct way: $\sqrt[3]{-27} = -3$, since $(-3)^3 = -27$. The memorizer carries the square-root rule ("negatives have no root") into cube roots, where it no longer applies. The two ideas sit so close together that the rule leaks across unless the reason (an odd number of negative factors stays negative) is spelled out.

Mistake 3: Assuming Double The Edge Means Double The Volume

Where it slips in: Scaling and model problems.

Don't do this: Claiming a box with edges twice as long holds twice as much.

The correct way: Volume scales with the cube of the length factor: $2^3 = 8$ times as much. Going the other way, doubling a volume multiplies each edge by $\sqrt[3]{2} \approx 1.26$.

Where Can You Get Help With Cubes And Cube Roots?

Cubes and cube roots usually click once a learner has built a few cubes from unit blocks and worked through prime factorisation by hand. If the patterns are still not landing, an algebra tutor or a middle school math tutor can walk through the factor-tree grouping one step at a time. For ongoing support, help with algebra and live math classes online cover exponents, roots and the algebra that follows.

Conclusion

• Cubes and cube roots are inverse operations: $a^3 = a \times a \times a$, and $\sqrt[3]{b}$ is the number whose cube is $b$.

• A perfect cube has every prime exponent in its factorisation divisible by 3, which is exactly what the factor-tree method checks.

• Non-perfect cubes such as 30 have irrational cube roots, found by estimating between neighbouring cubes and refining.

• Negative numbers have real, negative cube roots, the biggest difference from square roots.

• The most common errors are multiplying by 3 instead of cubing and forgetting that volume scales with the cube of length.

Practice These Before Moving On

1. Find $12^3$.

2. Find $\sqrt[3]{-216}$.

3. Find $\sqrt[3]{0.027}$.

4. Find the smallest number that 675 must be multiplied by to give a perfect cube, and name the cube root.

5. Find $\sqrt[3]{64 \times 125}$ using the product rule.

6. Estimate $\sqrt[3]{50}$ to two decimal places.

Answers:

1. $1728$

2. $-6$

3. $0.3$

4. $675 = 3^3 \times 5^2$, so multiply by 5 to get $3375 = 15^3$; the cube root is 15.

5. $4 \times 5 = 20$

6. About $3.68$ (it lies between 3 and 4, much closer to 4, since $64 - 50 = 14$ while $50 - 27 = 23$).

If question 4 stalls, go back to Method 1 and redraw the factor tree with the primes boxed in threes. Practice these to solidify your understanding, and if you would like a teacher to guide you, Book a free demo class.

Read More

• Cube root of 27

• Cube root of 64

• Cube root of 1331

• Perfect cube

• Sum of odd numbers

• Exponent rules