Find the cube roots of and .

Rational Numbers | FIO

Question 10

Find the cube roots of $27000$ and $10648$ .

Solution

Understand the Question

To find the cube root of a number using the prime factorisation method:
1. Find the prime factors of the number by continuous division until reaching $1$ .
2. Group identical prime factors into triplets (groups of three).
3. Take one factor from each triplet and multiply them together to obtain the cube root.

(i) Find the cube root of $27000$ .

Step 1 · Prime Factorisation of 27000

Dividing $27000$ by prime factors

\begin{aligned} 27000 \div 2 &= 13500 \\ 13500 \div 2 &= 6750 \\ 6750 \div 2 &= 3375 \\ 3375 \div 3 &= 1125 \\ 1125 \div 3 &= 375 \\ 375 \div 3 &= 125 \\ 125 \div 5 &= 25 \\ 25 \div 5 &= 5 \\ 5 \div 5 &= 1 \end{aligned}

Prime factorisation of $27000$ $27000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$

Step 2 · Calculate the Cube Root

Grouping the prime factors into triplets

\begin{aligned} \sqrt[3]{27000} &= \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3) \times (5 \times 5 \times 5)} \\[0.6em] &= 2 \times 3 \times 5 \\[0.6em] &= 30 \end{aligned}

Answer

(i) $30$

(ii) Find the cube root of $10648$ .

Step 1 · Prime Factorisation of 10648

Dividing $10648$ by prime factors

\begin{aligned} 10648 \div 2 &= 5324 \\ 5324 \div 2 &= 2662 \\ 2662 \div 2 &= 1331 \end{aligned}

Dividing $1331$ by $11$

\begin{aligned} 1331 \div 11 &= 121 \\ 121 \div 11 &= 11 \\ 11 \div 11 &= 1 \end{aligned}

Prime factorisation of $10648$ $10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11$

Step 2 · Calculate the Cube Root

Grouping the prime factors into triplets

\begin{aligned} \sqrt[3]{10648} &= \sqrt[3]{(2 \times 2 \times 2) \times (11 \times 11 \times 11)} \\[0.6em] &= 2 \times 11 \\[0.6em] &= 22 \end{aligned}

Answer

(ii) $22$

Common Mistakes

Pairs instead of Triplets: Grouping factors in pairs (like finding a square root) rather than in groups of three for a cube root.
Divisibility of 1331: Failing to test divisibility by $11$ after exhausting $2$ , $3$ , $5$ , and $7$ , since $1331 = 11^3$ .