Cube Root of 27 — Simplified Form, Methods, and Proof

The Answer At A Glance

Result: $\sqrt[3]{27} = 3$

Notation: $\sqrt[3]{27}$ or $27^{1/3}$.

Method shown: Prime factorisation, with cross-checks using direct verification and consecutive-cube bracketing.

Exact form: $3$ — an integer.

Type: $27$ is a perfect cube because $3^3 = 27$.

Quick Reference Table — Small Perfect Cubes

$27 = 3^3$ is the third entry in the perfect-cube sequence, after $1^3 = 1$ and $2^3 = 8$.

What "Cube Root of 27" Means

The cube root of a number $n$ is the value $x$ such that $x^3 = n$. For $\sqrt[3]{27}$, the $x$ with $x \cdot x \cdot x = 27$.

Verification: $3 \cdot 3 = 9$. $9 \cdot 3 = 27$. So $3^3 = 27$, and $\sqrt[3]{27} = 3$.

Unlike a square root, the cube root of a negative number is real: $\sqrt[3]{-27} = -3$.

How to Find ∛27 — Three Methods

Method 1 — Prime factorisation

Factor $27$ into primes:

$$27 = 3 \cdot 9 = 3 \cdot 3 \cdot 3 = 3^3$$

Group the prime factors in triples:

$$\sqrt[3]{27} = \sqrt[3]{3^3} = 3$$

Method 2 — Direct verification

$$3^3 = 3 \cdot 3 \cdot 3 = 9 \cdot 3 = 27 \checkmark$$

Method 3 — Bracketing by consecutive cubes

$$2^3 = 8, \quad 3^3 = 27, \quad 4^3 = 64$$

$27 = 3^3$ exactly. So $\sqrt[3]{27} = 3$.

Is ∛27 Rational or Irrational?

$\sqrt[3]{27}$ is rational — in fact, an integer. $27$ is a perfect cube ($27 = 3^3$), and the cube root of a perfect cube is always a whole number.

Compare with $\sqrt[3]{28}$ or $\sqrt[3]{26}$ — both irrational.

A Proof Using The Algebraic Identity

For a quick algebraic confirmation, use the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Setting $a^3 = 27$ and $b = 3$ (so $b^3 = 27$):

$$27 - 27 = (a - 3)(a^2 + 3a + 9)$$

The left side is zero. For the right side to be zero, $a - 3 = 0$ (the other factor $a^2 + 3a + 9$ has discriminant $9 - 36 = -27 < 0$, no real roots).

So $a = 3$, confirming $\sqrt[3]{27} = 3$.

Where ∛27 Shows Up

$\sqrt[3]{27}$ comes up whenever you invert a cubic relationship.

• Cube volumes. A cube with volume $27$ cubic units has a side length of $3$ units.

• Scaling solids. Tripling the linear dimensions of a 3D shape scales the volume by $3^3 = 27$. Reversing — going from a volume increase of $27$ to the linear scale of $3$ — uses $\sqrt[3]{27}$.

• Polynomial roots. The equation $x^3 = 27$ has the real solution $x = 3$, and two complex solutions involving $\sqrt{-3}$. The real cube root is the integer $3$.

• Sum of cubes. $27 + 1 = 28$ is not a perfect cube. But $27 = 3^3$ appears in identities like $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ — setting $a = 3$, $b = 0$ recovers $27$.

Tripping Points To Avoid On Cube-Root Problems

Mistake 1: Confusing the cube root with the square root.

Where it slips in: Reading $\sqrt[3]{27}$ but reaching for the square-root key.

Don't do this: $\sqrt[3]{27} = \sqrt{27} \approx 5.196$.

The correct way: The small $3$ above the radical means cube root. $\sqrt[3]{27} = 3$ (integer). $\sqrt{27} = 3\sqrt{3} \approx 5.196$ (irrational). Different operations.

In Bhanzu's Grade 7 cohorts, the cube-root-vs-square-root confusion shows up on roughly three out of ten first attempts. A Bhanzu trainer who sees a student writing $\sqrt{27}$ when the problem asks for $\sqrt[3]{27}$ points to the small $3$ in red, draws an arrow to it, and the index lands in memory by the next problem.

Mistake 2: Squaring instead of cubing during verification.

Where it slips in: Verifying $\sqrt[3]{27} = 3$ by computing $3^2$ instead of $3^3$.

Don't do this: $3^2 = 9 \neq 27$, so $\sqrt[3]{27} \neq 3$.

The correct way: Cube root means cube to verify. $3^3 = 27 \checkmark$.

Mistake 3: Forgetting the cube root of a negative is well-defined.

Where it slips in: Refusing to compute $\sqrt[3]{-27}$ on the grounds that "roots of negatives are not real."

Don't do this: $\sqrt[3]{-27}$ → undefined.

The correct way: $\sqrt[3]{-27} = -3$. The cube root of a negative number is the negative of the cube root of the positive number. Only even roots of negatives leave the reals.

Conclusion

• The cube root of 27 is exactly $3$, because $3^3 = 27$.

• $27$ is a perfect cube — its cube root is a clean integer.

• Three methods confirm the value: prime factorisation, direct verification, and bracketing between consecutive cubes.

• The cube root of a negative number is well-defined in the reals: $\sqrt[3]{-27} = -3$.

• Memorising the cubes from $1^3$ to $10^3$ turns most cube-root questions into a single-step lookup.

A practical next step

1. Find $\sqrt[3]{125}$ using prime factorisation.

2. A cube has volume $27$ cubic metres. What is the length of each side?

3. Compute $\sqrt[3]{-27}$ and verify by cubing the result.

Want a Bhanzu trainer to walk you through more cube-root problems? Book a free demo class — online globally.