Cubes and Cube Roots

Cubes, perfect cubes and cube roots of whole numbers, found by prime factorisation

Before you start

You have found a square root by halving each index in the prime factorisation: $\sqrt{2^{4} \times 3^{2}} = 2^{2} \times 3 = 12$ . Cube roots work the same way, with 3 in place of 2.

Cubes and Perfect Cubes

When three of the same number are multiplied together, the result is the cube of that number: $5 \times 5 \times 5 = 5^{3} = 125$ . We read $5^{3}$ as "5 cubed", and 125 is the cube of 5.

Definition:

The cube of a number $n$ is $n \times n \times n$ , written $n^{3}$ . A perfect cube is the cube of a whole number.

The cubes of 1 to 10 are worth knowing by heart:

$n$	1	2	3	4	5	6	7	8	9	10
$n^{3}$	1	8	27	64	125	216	343	512	729	1000

Example 1: A cube in index notation

Given that $12 = 2^{2} \times 3$ , write $12^{3}$ as a product of its prime factors in index notation.

Step 1: Write the prime factorisation of 12 three times, because $12^{3} = 12 \times 12 \times 12$ .

$12^{3} = (2^{2} \times 3) \times (2^{2} \times 3) \times (2^{2} \times 3)$

Step 2: Count each prime factor: there are $2 + 2 + 2 = 6$ twos and $1 + 1 + 1 = 3$ threes.

$12^{3} = 2^{6} \times 3^{3}$

Answer: $12^{3} = 2^{6} \times 3^{3}$ . Cubing a number multiplies each index in its prime factorisation by 3.

Cube Roots

Finding a cube root is the reverse of cubing. Since $5^{3} = 125$ , the cube root of 125 is 5. We write $\sqrt[3]{125} = 5$ .

Definition:

The cube root of a number is the number whose cube is that number. If $a^{3} = b$ , then $\sqrt[3]{b} = a$ .

The small 3 is part of the sign. Without it, the sign means a square root.

$2^{3} = 8$ , so $\sqrt[3]{8} = 2$

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

$4^{3} = 64$ , so $\sqrt[3]{64} = 4$

Cubes and cube roots in a cube

A cube with edges of length $n$ has volume $n \times n \times n = n^{3}$ . So if we know the volume of a cube, the length of an edge is $\sqrt[3]{\text{volume}}$ .

Perfect Cubes and Cube Roots by Prime Factorisation

If the prime factors of a number can be shared into three equal groups, the number is a perfect cube and one group is its cube root. For this, each prime factor must appear 3, 6, 9, ... times.

Rule: Perfect cubes and their cube roots

A whole number greater than 1 is a perfect cube if every index in its prime factorisation is a multiple of 3. If any index is not a multiple of 3, it is not a perfect cube.
To find the cube root of a perfect cube, divide each index by 3.

Example 2: A cube root by prime factorisation

Find $\sqrt[3]{216}$ by prime factorisation.

Step 1: Write 216 as a product of its prime factors, dividing by prime numbers in turn.

$216 = 2^{3} \times 3^{3}$

Step 2: Both indices are 3, so the prime factors can be shared into three equal groups.

$216 = (2 \times 3) \times (2 \times 3) \times (2 \times 3) = (2 \times 3)^{3}$

Step 3: One group is the cube root. This is the same as dividing each index by 3.

$\sqrt[3]{216} = 2 \times 3 = 6$

Answer: $\sqrt[3]{216} = 6$ . Check: $6 \times 6 \times 6 = 216$ .

Example 3: Is it a perfect cube?

Decide whether each number is a perfect cube. If it is, find its cube root.

(a) 2744

(b) $2^{6} \times 5^{2}$

(a) Write 2744 as a product of its prime factors.

$2744 = 2^{3} \times 7^{3}$

Each index is 3, a multiple of 3, so 2744 is a perfect cube.

Divide each index by 3: $\sqrt[3]{2744} = 2 \times 7 = 14$

(b) The index of 2 is 6, a multiple of 3, but the index of 5 is 2, which is not. Two 5s cannot be shared into three equal groups.

Answer: (a) 2744 is a perfect cube and $\sqrt[3]{2744} = 14$ . (b) $2^{6} \times 5^{2}$ is not a perfect cube.

Making a Perfect Cube

If a number is not a perfect cube, we can multiply or divide it so that every index becomes a multiple of 3. Multiplying raises an index; dividing lowers it. When dividing, a prime factor whose index is 1 or 2 is already below 3, so it has to be divided away completely.

Example 4: Making a perfect cube

Find the smallest positive whole number that 432 must be

(a) multiplied by to give a perfect cube,

(b) divided by to give a perfect cube.

Write 432 as a product of its prime factors.

$432 = 2^{4} \times 3^{3}$

The index of 3 is already a multiple of 3. The index of 2 is 4, which is not.

(a) The next multiple of 3 after 4 is 6, so two more 2s are needed. Multiply by $2^{2} = 4$ .

$432 \times 4 = 2^{6} \times 3^{3} = 1728$

(b) The multiple of 3 just below 4 is 3, so one 2 must go. Divide by 2.

$432 \div 2 = 2^{3} \times 3^{3} = 216$

Answer: (a) The smallest number to multiply by is 4, which gives $1728 = 12^{3}$ . (b) The smallest number to divide by is 2, which gives $216 = 6^{3}$ .

Cube Roots in Word Problems

Problems about the volume of a cube often need a cube root.

Example 5: The edge of a cube

The volume of a cube is $1728 \text{ cm}^{3}$ . Find the length of an edge of the cube.

Step 1: Volume of a cube = edge × edge × edge, so the length of an edge is the cube root of the volume.

Length of an edge $= \sqrt[3]{\text{volume}} = \sqrt[3]{1728}$

Step 2: Write 1728 as a product of its prime factors.

$1728 = 2^{6} \times 3^{3}$

Step 3: Share the six 2s and the three 3s into three equal groups. This divides each index by 3.

$1728 = (2^{2} \times 3) \times (2^{2} \times 3) \times (2^{2} \times 3)$

$\sqrt[3]{1728} = 2^{2} \times 3 = 12$

Answer: The length of an edge of the cube is 12 cm. Check: $12 \times 12 \times 12 = 1728$ .

Both a Perfect Square and a Perfect Cube

Some numbers are both: $64 = 2^{6}$ is $8^{2}$ and also $4^{3}$ . A perfect square has every index even and a perfect cube has every index a multiple of 3, so a number that is both has every index a multiple of 6.

Example 6: Two roots of one number

Show that 729 is both a perfect square and a perfect cube. Find $\sqrt{729}$ and $\sqrt[3]{729}$ .

Step 1: Write 729 as a product of its prime factors.

$729 = 3^{6}$

Step 2: The index 6 is even, so 729 is a perfect square. Halve the index.

$\sqrt{729} = 3^{3} = 27$

Step 3: The index 6 is also a multiple of 3, so 729 is a perfect cube. Divide the index by 3.

$\sqrt[3]{729} = 3^{2} = 9$

Answer: $\sqrt{729} = 27$ and $\sqrt[3]{729} = 9$ , so 729 is both $27^{2}$ and $9^{3}$ .

Common Mistakes

Cubing is not multiplying by 3. $2^{3} = 2 \times 2 \times 2 = 8$ , not 6. In the same way, $\sqrt[3]{27} = 3$ , not 9.
For a cube root, divide each index by 3, not by 2. $\sqrt[3]{2^{6}} = 2^{2}$ , not $2^{3}$ .
The edge of a cube comes from its volume, not from an area. The length of an edge is $\sqrt[3]{\text{volume}}$ .

Practice

Work each question on paper before you open its solution.

Question 1

Find the value of

(a) $9^{3}$

(b) $\sqrt[3]{343}$

ShowHide solution

(a) $9^{3} = 9 \times 9 \times 9 = 81 \times 9 = 729$

(b) $7 \times 7 \times 7 = 49 \times 7 = 343$ , so $\sqrt[3]{343} = 7$

Answer: (a) $9^{3} = 729$ (b) $\sqrt[3]{343} = 7$

Question 2

Given that $18 = 2 \times 3^{2}$ , write $18^{3}$ as a product of its prime factors in index notation.

ShowHide solution

Write the prime factorisation of 18 three times.

$18^{3} = (2 \times 3^{2}) \times (2 \times 3^{2}) \times (2 \times 3^{2})$

There are three 2s and $2 + 2 + 2 = 6$ threes, so each index has been multiplied by 3.

Answer: $18^{3} = 2^{3} \times 3^{6}$

Question 3

Find $\sqrt[3]{3375}$ by prime factorisation.

ShowHide solution

Write 3375 as a product of its prime factors.

$3375 = 3^{3} \times 5^{3} = (3 \times 5) \times (3 \times 5) \times (3 \times 5)$

Both indices are multiples of 3. Divide each index by 3.

$\sqrt[3]{3375} = 3 \times 5 = 15$

Answer: $\sqrt[3]{3375} = 15$

Question 4

State whether each number is a perfect cube. If it is, find its cube root.

(a) $2^{9} \times 7^{3}$

(b) $3^{6} \times 5^{4}$

ShowHide solution

(a) The indices 9 and 3 are both multiples of 3, so the number is a perfect cube. Divide each index by 3.

$\sqrt[3]{2^{9} \times 7^{3}} = 2^{3} \times 7 = 8 \times 7 = 56$

(b) The index of 5 is 4, which is not a multiple of 3, so the number is not a perfect cube.

Answer: (a) A perfect cube; its cube root is 56. (b) Not a perfect cube.

Question 5

Find the smallest positive whole number that 1080 must be multiplied by to give a perfect cube. What is the cube root of that perfect cube?

ShowHide solution

Write 1080 as a product of its prime factors.

$1080 = 2^{3} \times 3^{3} \times 5$

The indices of 2 and 3 are already 3. There is only one 5, so two more 5s are needed. Multiply by $5^{2} = 25$ .

$1080 \times 25 = 2^{3} \times 3^{3} \times 5^{3} = 27\,000$

$\sqrt[3]{27\,000} = 2 \times 3 \times 5 = 30$

Answer: The smallest number to multiply by is 25. The perfect cube is 27 000 and its cube root is 30.

Question 6

Given that $6048 = 2^{5} \times 3^{3} \times 7$ , find the smallest positive whole number that 6048 must be divided by to give a perfect cube.

ShowHide solution

Dividing can only lower an index, so each index must come down to a multiple of 3. The index of 2 comes down from 5 to 3, so $2^{2}$ must go. The index of 3 stays at 3. The index of 7 is 1, which is less than 3, so the 7 must be divided away completely.

Divide by $2^{2} \times 7 = 28$ .

$6048 \div 28 = 2^{3} \times 3^{3} = 216 = 6^{3}$

Answer: The smallest number to divide by is 28.

Question 7

A cubical tank holds exactly 64 litres of water when it is full. Find the length, in centimetres, of an inside edge of the tank. (1 litre = $1000 \text{ cm}^{3}$ )

ShowHide solution

Change the volume to cubic centimetres.

Volume $= 64 \times 1000 = 64\,000 \text{ cm}^{3}$

Since $64 = 2^{6}$ and $1000 = 2^{3} \times 5^{3}$ , there are $6 + 3 = 9$ twos and three 5s.

$64\,000 = 2^{9} \times 5^{3}$

The edge is the cube root of the volume, so divide each index by 3.

$\sqrt[3]{64\,000} = 2^{3} \times 5 = 8 \times 5 = 40$

Answer: The length of an inside edge of the tank is 40 cm.

Question 8

Ali fills a cubical box exactly with 9261 small cubes, each of edge 1 cm.

(a) How many small cubes lie along one edge of the box?

(b) How many small cubes are in the bottom layer?

ShowHide solution

(a) The small cubes fill a volume of $9261 \text{ cm}^{3}$ , so an edge of the box is $\sqrt[3]{9261}$ cm long. Each small cube is 1 cm wide, so the same number of small cubes lie along one edge.

$9261 = 3^{3} \times 7^{3}$

$\sqrt[3]{9261} = 3 \times 7 = 21$

(b) The bottom layer is a square with 21 cubes along each side.

$21 \times 21 = 441$

Answer: (a) 21 small cubes (b) 441 small cubes

Question 9

A number is written as $2^{6} \times 5^{6}$ .

(a) Explain why the number is both a perfect square and a perfect cube.

(b) Find its square root and its cube root.

ShowHide solution

(a) Both indices are 6. Since 6 is even, the number is a perfect square. Since 6 is a multiple of 3, it is also a perfect cube.

(b) Halve each index for the square root; divide each index by 3 for the cube root.

$\sqrt{2^{6} \times 5^{6}} = 2^{3} \times 5^{3} = 8 \times 125 = 1000$

$\sqrt[3]{2^{6} \times 5^{6}} = 2^{2} \times 5^{2} = 4 \times 25 = 100$

Answer: The square root is 1000 and the cube root is 100.

Key Points

The cube of $n$ is $n^{3} = n \times n \times n$ . A perfect cube is the cube of a whole number.
The cube root undoes cubing: if $a^{3} = b$ , then $\sqrt[3]{b} = a$ .
Cubing a number multiplies each index in its prime factorisation by 3.
A whole number greater than 1 is a perfect cube if every index in its prime factorisation is a multiple of 3, and is not one if any index is not. Its cube root is found by dividing each index by 3.
To make a perfect cube, multiply to raise each index to the next multiple of 3, or divide to lower it to the multiple of 3 below (a prime factor with index 1 or 2 is divided away completely).
The length of an edge of a cube is $\sqrt[3]{\text{volume}}$ .
A number that is both a perfect square and a perfect cube has every index a multiple of 6.