Find cube root of 13824 by prime factorisation

Question

Find the cube root of 13824 by prime factorisation.

Solution — Step by Step

We start dividing 13824 by the smallest prime (2) and keep going:

13824 \div 2 = 6912

6912 \div 2 = 3456

3456 \div 2 = 1728

1728 \div 2 = 864

864 \div 2 = 432

432 \div 2 = 216

216 \div 2 = 108

108 \div 2 = 54

54 \div 2 = 27

27 \div 3 = 9

9 \div 3 = 3

3 \div 3 = 1

13824 = 2^9 \times 3^3

Let’s verify: $2^9 = 512$ and $3^3 = 27$ , so $512 \times 27 = 13824$ . ✓

For a perfect cube, every prime factor must appear a multiple of 3 times.

13824 = \underbrace{2 \times 2 \times 2}_{\text{group 1}} \times \underbrace{2 \times 2 \times 2}_{\text{group 2}} \times \underbrace{2 \times 2 \times 2}_{\text{group 3}} \times \underbrace{3 \times 3 \times 3}_{\text{group 4}}

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

Check: $24^3 = 24 \times 24 \times 24 = 576 \times 24 = 13824$ ✓

Why This Works

A perfect cube has each prime factor appearing exactly 3 times (or a multiple of 3 times). When we take the cube root, we pick one from each group of three identical factors. This is because $\sqrt[3]{a^3} = a$ .

So $\sqrt[3]{2^9 \times 3^3} = 2^{9/3} \times 3^{3/3} = 2^3 \times 3^1 = 8 \times 3 = 24$ .

Quick check before you start: the digit sum of 13824 is $1+3+8+2+4 = 18$ , which is divisible by 9 (hence by 3). And the number is even, so both 2 and 3 are factors. This narrows down your factorisation quickly.

Alternative Method

Recognise that $13824 = 8 \times 1728 = 8 \times 12^3$ ? Actually, let’s check: $12^3 = 1728$ and $8 \times 1728 = 13824$ . So:

\sqrt[3]{13824} = \sqrt[3]{8 \times 1728} = \sqrt[3]{8} \times \sqrt[3]{1728} = 2 \times 12 = 24

This shortcut works when you recognise a factor as a perfect cube.

Common Mistake

Students sometimes stop the division tree early and miss factors. Always divide until you reach 1. Also, when grouping into triples, students sometimes take one factor from each “pair” instead of each “triple.” For square roots we take pairs; for cube roots we must form groups of three.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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