Cube Root of 6 - Value, Is It Irrational? Examples

What Is A Cube Root?

The cube root of a number $n$ is the value $r$ such that $r^3 = n$, the number that, multiplied by itself three times, gives $n$. The cube root of 6 is the number whose cube is 6.

No whole number works here: $1^3 = 1$ (too small) and $2^3 = 8$ (too big). So $\sqrt[3]{6}$ lies between 1 and 2, and because 6 is much closer to 8 than to 1, the value lands near the top of that gap, at about $1.817$.

A cube root is not the same as dividing by 3. Dividing 6 by 3 gives 2, but $2^3 = 8$, not 6, and the cube root undoes three equal factors, not a single multiplication.

Where ∛6 Appears In Real Math

$\sqrt[3]{6}$ is the edge length of a cube that holds exactly 6 cubic units of volume - since a cube of edge $s$ has volume $s^3$, solving $s^3 = 6$ gives $s = \sqrt[3]{6} \approx 1.817$ units. It also appears whenever a volume, mass, or scaling factor triples down to a single length: a tank built to hold six times a unit cube is only about 1.8 times wider, not six times, which is why doubling a recipe rarely doubles the pot size.

Quick Reference Table

Is The Cube Root Of 6 Rational Or Irrational?

$\sqrt[3]{6}$ is irrational - it cannot be written as a fraction $\frac{p}{q}$ of two integers, and its decimal neither terminates nor repeats.

Quick reasoning. A whole number has a rational cube root only when it is a perfect cube, like 1, 8, 27, or 64. The prime factorisation of 6 is $2 \times 3$, and no prime appears three times, so 6 is not a perfect cube and $\sqrt[3]{6}$ is irrational.

A short proof by contradiction. Suppose $\sqrt[3]{6}$ were rational. Then we could write it in lowest terms as $\sqrt[3]{6} = \frac{p}{q}$ with no common factor. Cubing both sides:

$$6 = \frac{p^3}{q^3}$$

$$p^3 = 6 q^3 = 2 \times 3 \times q^3$$

So $p^3$ is even, which forces $p$ to be even; write $p = 2k$. Then:

$$8k^3 = 6q^3$$

$$4k^3 = 3q^3$$

The left side is even, so $3q^3$ is even, so $q^3$ is even, so $q$ is even. Now $p$ and $q$ are both even, contradicting the lowest-terms assumption. No such fraction exists, so $\sqrt[3]{6}$ is irrational. The same argument retires every non-perfect-cube integer to the set of irrational numbers.

What this means in practice. Any decimal you write for $\sqrt[3]{6}$ is an approximation; the exact value lives only in the symbol $\sqrt[3]{6}$. Keep the radical form in algebra, and round only at the final step.

How Do You Find ∛6? (Bracketing And Prime Factorisation)

There is no long-division shortcut for cube roots the way there is for square roots, so you bracket the value between perfect cubes and then refine.

Method 1: Prime factorisation (to check it cannot simplify).

Write 6 as a product of primes.

$$6 = 2 \times 3$$

A cube root simplifies only when a prime appears three times, so a group of three can leave the radical. Here 2 and 3 each appear once, so nothing escapes:

$$\sqrt[3]{6} = \sqrt[3]{2 \times 3}$$

Final answer: $\sqrt[3]{6}$ is already in simplest form.

Method 2: Bracket, then refine.

Start by trapping the value between consecutive cubes.

$$1^3 = 1 \le 6 \le 8 = 2^3$$

So $1 < \sqrt[3]{6} < 2$. Test a middle guess.

$$1.8^3 = 5.832 \ (\text{a little low})$$

$$1.82^3 = 6.028 \ (\text{a little high})$$

Refine once with a Newton step, $x_{\text{new}} = x - \dfrac{x^3 - 6}{3x^2}$, starting at $x = 1.8$.

$$x_{\text{new}} = 1.8 - \frac{5.832 - 6}{3 \times 3.24}$$

$$x_{\text{new}} = 1.8 + \frac{0.168}{9.72}$$

$$x_{\text{new}} \approx 1.8171$$

Final answer: $\sqrt[3]{6} \approx 1.817$. The process never lands on a clean decimal, which is exactly what "irrational" means.

Examples Of Cube Root Of 6

Example 1

Evaluate $(\sqrt[3]{6})^3$.

Cubing undoes the cube root by definition.

$$(\sqrt[3]{6})^3 = 6$$

Final answer: $6$.

Example 2

A common slip worth walking through: find $\sqrt[3]{6}$ by "dividing off" the root.

The tempting first move is to treat the cube root like a division and write $\sqrt[3]{6} = \frac{6}{3} = 2$.

Check it.

$$2^3 = 8$$

But $8 \ne 6$, so 2 is too big — the answer must be less than 2. A cube root asks "what number cubed gives 6?", not "6 shared into 3 parts".

Bracket instead: $1^3 = 1$ and $2^3 = 8$, so the value sits between 1 and 2, near $1.817$.

Final answer: $\sqrt[3]{6} \approx 1.817$, not 2.

Example 3

Simplify $\sqrt[3]{48}$ and spot where $\sqrt[3]{6}$ hides inside it.

Factor out the largest perfect cube.

$$48 = 8 \times 6$$

$$\sqrt[3]{48} = \sqrt[3]{8} \times \sqrt[3]{6}$$

$$\sqrt[3]{48} = 2\sqrt[3]{6}$$

The first instinct here is to reach for the square-root reflex and pull out a pair; a cube root only releases a factor when three copies group together, so $8 = 2^3$ leaves a clean 2 outside.

Final answer: $\sqrt[3]{48} = 2\sqrt[3]{6} \approx 3.634$.

Example 4

Solve $x^3 = 6$ for the real value of $x$.

Take the cube root of both sides.

$$x = \sqrt[3]{6}$$

$$x \approx 1.817$$

Unlike a square root, a cube root of a positive number has exactly one real value, so there is no $\pm$ here.

Final answer: $x = \sqrt[3]{6} \approx 1.817$.

Common Mistakes

Mistake 1: Treating the cube root as "divide by 3"

Where it slips in: The very first attempt at $\sqrt[3]{6}$, especially for a student who just met the notation.

Don't do this: Writing $\sqrt[3]{6} = 6 \div 3 = 2$.

The correct way: Ask what number cubed gives 6. Since $1^3 = 1$ and $2^3 = 8$, the answer sits between 1 and 2, at about $1.817$. The confusion between the cube-root sign and a divide-by-three step is the single most common source of wrong answers here, and it clears up the moment you cube your guess to check it.

Mistake 2: Thinking ∛6 can be simplified

Where it slips in: Students assume every radical breaks down into something neater.

Don't do this: Writing $\sqrt[3]{6} = \sqrt[3]{2} \times \sqrt[3]{3}$ and calling it "simpler".

The correct way: That split is true, but neither piece leaves the radical, so it is not simpler. A cube root simplifies only when a prime appears three times; in $6 = 2 \times 3$ nothing does, so $\sqrt[3]{6}$ is already the simplest form.

Mistake 3: Rounding too early

Where it slips in: Multi-step problems where $\sqrt[3]{6}$ appears in the middle of a calculation.

Don't do this: Replace $\sqrt[3]{6}$ with $1.817$ at the start and carry that rounded value through every step.

The correct way: Keep $\sqrt[3]{6}$ in radical form until the final line, then round once. Early rounding compounds error across multiplications, and in a case like $\sqrt[3]{48} = 2\sqrt[3]{6}$ the exact form is also shorter to write.

Conclusion

The cube root of 6 is $\sqrt[3]{6} \approx 1.817$: irrational, already in simplest form, and best understood as the edge of a cube that holds 6 cubic units. Bracket it between $1^3$ and $2^3$, refine with one Newton step, and keep the radical until the final answer. To take cube roots and radicals further with a teacher, explore Bhanzu's algebra tutor, a high school math tutor, or live math classes online. Prefer to see the method taught live? Book a free demo class.

Read More

• Cube Root of 3 — the value and irrationality of another non-perfect cube

• Cube Root of 5 — a close neighbour worked the same way

• Cube Root of 24 — where a perfect-cube factor does come out

• Square Root of 10 — the square-root version of an irrational root

• Simplifying Radical Expressions — the full method for pulling factors out of any radical

• Square Root Tricks — fast estimation methods for roots without a calculator