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

TL;DR

The cube root of 27 is exactly 3, because 3×3×3=27. 27 is a perfect cube, so the cube root is a clean integer — no decimal approximation needed. This article covers the value, the prime-factorisation method, where the cube root shows up, and the slips students make most often.

The Answer At A Glance

Result: 3

Notation: (27^{1/3}) or (3\sqrt[3]{27}

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=3³ is the third entry in the perfect-cube sequence, after 1 and 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,) confirming (\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 = 27✓)

Method 3 — Bracketing by consecutive cubes

(2^3=8,3^3=27,4^3=64)

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

Is ∛27 Rational or Irrational?

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

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):

(27−27=(a−3)(a^2+3a+9)).

The left side is zero. For the right side to be zero, (a−3=0) confirming (\sqrt[3]{27}=3).

Where ∛27 Shows Up

(\sqrt[3]{27}) comes up whenever you find cubic relationships.

• 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 27.
• Polynomial roots. The equation (x^3=27) has the real solution x=3.
• Sum of cubes. 27+1=28 is not a perfect cube.

Tripping Points To Avoid On Cube-Root Problems

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

• Where it slips: Reading (\sqrt[3]{27}) but reaching for the square-root key.
• The correct way: (\sqrt[3]{27} = 3) is integer, while (\sqrt{27} \approx 5.196).

Mistake 2: Squaring instead of cubing during verification.

• Where it slips: Verifying (3) by computing (3^2) instead of (3^3).
• The correct way: Verify by cubing: (3^3 = 27✓).

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

• Where it slips: Refusing to compute (-\sqrt[3]{-27}) on the grounds that "roots of negatives are not real."
• The correct way: (-\sqrt[3]{-27} = -3).

Conclusion

• The cube root of 27 is exactly 3.
• 27 is a perfect cube and its cube root is a clean integer.
• Three methods confirm the value: prime factorisation, direct verification, and consecutive cubes.
• The cube root of a negative number is well-defined: (-\sqrt[3]{-27} = -3).

Frequently Asked Questions

What is the cube root of 27? (\sqrt[3]{27} = 3).

Is 27 a perfect cube? Yes, because (27=3^3).

How do you find the cube root of 27? Prime-factorise: (27=3^3), and the cube root is 3.

What is (-\sqrt[3]{-27})? (-\sqrt[3]{-27} = -3).

Is the cube root of 27 rational or irrational? Rational — it is an integer.