Cube roots

You know that $6^3 = 216$. Turn the ques­tion round: which num­ber, mul­ti­plied by itself three times, gives $216$? The answer, $6$, is the cube root of $216$. Find­ing a cube root is the oppo­site of cub­ing. In this les­son you will meet the cube root sign, find cube roots by group­ing prime fac­tors in threes, and learn a quick way to find the cube root of a per­fect cube from its last digit.

What a cube root is#

If $a^3 = b$, then $a$ is called the cube root of $b$. We write it with the sign $\sqrt[3]{\phantom{b}}$, like this:

$$a^3 = b \quad \Rightarrow \quad \sqrt[3]{b} = a$$

We read $\sqrt[3]{b}$ as "the cube root of b". The small $3$ in the sign tells it apart from a square root.

$$\begin{aligned} 2^3 = 8 &\quad \Rightarrow \quad \sqrt[3]{8} = 2 \\ 3^3 = 27 &\quad \Rightarrow \quad \sqrt[3]{27} = 3 \\ 10^3 = 1000 &\quad \Rightarrow \quad \sqrt[3]{1000} = 10 \end{aligned}$$

From the table of cubes you can read many cube roots straight away: $\sqrt[3]{64} = 4$, $\sqrt[3]{343} = 7$, $\sqrt[3]{729} = 9$.

Cube roots by prime fac­tori­sa­tion#

1. Break the num­ber into prime fac­tors.
2. Group equal prime fac­tors in threes.
3. Take one prime from each group of three.
4. Mul­ti­ply them. The answer is the cube root.

Each group of three, such as $2 \times 2 \times 2$, is a cube. Tak­ing one from the group is tak­ing its cube root.

Exam­ple 1#

Find $\sqrt[3]{216}$.

$$\begin{aligned} \sqrt[3]{216} &= \sqrt[3]{(2 \times 2 \times 2) \times (3 \times 3 \times 3)} \\ &= 2 \times 3 \\ &= 6 \end{aligned}$$

Exam­ple 2#

Find $\sqrt[3]{2744}$.

Divide by $2$ three times: $2744$, $1372$, $686$, $343$. Then $343 = 7 \times 7 \times 7$.

$$\begin{aligned} \sqrt[3]{2744} &= \sqrt[3]{(2 \times 2 \times 2) \times (7 \times 7 \times 7)} \\ &= 2 \times 7 \\ &= 14 \end{aligned}$$

Exam­ple 3#

Find $\sqrt[3]{35937}$.

The dig­its add to $3 + 5 + 9 + 3 + 7 = 27$, so $3$ divides it. Divid­ing by $3$ three times: $35937$, $11979$, $3993$, $1331$. And $1331 = 11 \times 11 \times 11$.

$$\begin{aligned} \sqrt[3]{35937} &= \sqrt[3]{(3 \times 3 \times 3) \times (11 \times 11 \times 11)} \\ &= 3 \times 11 \\ &= 33 \end{aligned}$$

Exam­ple 4#

Find $\sqrt[3]{13824}$.

Divid­ing by $2$ nine times gives $27$, and $27 = 3 \times 3 \times 3$. Nine 2s make three groups of three.

$$\begin{aligned} \sqrt[3]{13824} &= \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (3 \times 3 \times 3)} \\ &= 2 \times 2 \times 2 \times 3 \\ &= 24 \end{aligned}$$

Is it a per­fect cube?#

If every prime can be put in a group of three, the num­ber is a per­fect cube. If some primes are left over, it is not.

Exam­ple 5#

Is $500$ a per­fect cube? If not, find the small­est num­ber you must mul­ti­ply it by to make one.

$$500 = 2 \times 2 \times (5 \times 5 \times 5)$$

The two 2s do not make a group of three. One more $2$ would com­plete the group. So mul­ti­ply by $2$.

$$500 \times 2 = 1000 = (2 \times 2 \times 2) \times (5 \times 5 \times 5)$$

Now $\sqrt[3]{1000} = 2 \times 5 = 10$.

Exam­ple 6#

Find the small­est num­ber by which $72$ must be divided to give a per­fect cube.

$$72 = (2 \times 2 \times 2) \times 3 \times 3$$

The two 3s are left over. Divide them away: divide by $3 \times 3 = 9$.

$$72 \div 9 = 8 = 2 \times 2 \times 2$$

So divide by $9$, and the per­fect cube is $8$.

Exam­ple 7#

Find the small­est num­ber by which $1350$ must be mul­ti­plied to give a per­fect cube.

$$1350 = 2 \times (3 \times 3 \times 3) \times 5 \times 5$$

The $2$ needs two more 2s. The two 5s need one more $5$. So mul­ti­ply by $2 \times 2 \times 5 = 20$.

$$1350 \times 20 = 27000 = 30^3$$

A quick way for per­fect cubes#

If you know a num­ber is a per­fect cube with up to six dig­its, you can find its cube root in your head. Use the last-digit table from the first les­son.

1. Mark off the last three dig­its. The last digit of the num­ber gives the last digit of the root.
2. Look at the dig­its left in front. Find the biggest cube that is not more than them. Its cube root is the first digit of the answer.

Exam­ple 8#

Find $\sqrt[3]{13824}$ the quick way.

Mark off: $13 \mid 824$. The num­ber ends in $4$, so the root ends in $4$.

In front we have $13$. Now $2^3 = 8$ is not more than $13$, but $3^3 = 27$ is. So the first digit is $2$.

$$\sqrt[3]{13824} = 24$$

This agrees with Exam­ple 4.

Exam­ple 9#

Find $\sqrt[3]{110592}$, given that it is a per­fect cube.

Mark off: $110 \mid 592$. The num­ber ends in $2$, so the root ends in $8$. In front we have $110$: $4^3 = 64$ is not more than $110$, but $5^3 = 125$ is. So the first digit is $4$.

$$\sqrt[3]{110592} = 48$$

This quick way only works for per­fect cubes. For any other num­ber, use prime fac­tori­sa­tion first to check.

Remem­ber. Group equal prime fac­tors in threes and take one from each group. Primes left over mean the num­ber is not a per­fect cube.

Com­mon mis­takes#

• Group­ing in pairs, as for square roots. Cube roots need groups of three.
• Tak­ing all three primes from a group. Take only one.
• Using the quick way on a num­ber that is not a per­fect cube. It will give a wrong answer.
• Mix­ing up the last dig­its $2$ and $8$, or $3$ and $7$. They swap.

Prac­tice#

1. Find $\sqrt[3]{3375}$ by prime fac­tori­sa­tion.
2. Find $\sqrt[3]{1728}$.
3. Find $\sqrt[3]{9261}$.
4. Find $\sqrt[3]{10648}$.
5. Find $\sqrt[3]{74088}$.
6. Find the small­est num­ber by which $675$ must be mul­ti­plied to give a per­fect cube.
7. Find the small­est num­ber by which $128$ must be divided to give a per­fect cube.
8. Find the small­est num­ber by which $3087$ must be divided to give a per­fect cube, and find the cube root of the result.
9. Given that $175616$ is a per­fect cube, find its cube root the quick way.
10. Given that $857375$ is a per­fect cube, find its cube root the quick way.

Answers#