Cubing uses a number as a factor three times: $4^{3} = 4 \cdot 4 \cdot 4 = 64$. A cube root undoes cubing, the way a square root undoes squaring.
Why it works
A cube with edges 4 units long is built from 4 layers of $4 \times 4$ unit cubes: $4 \cdot 4 \cdot 4 = 64$ cubes. So if a cube holds 64 unit cubes, its edge is 4. We write $\sqrt[3]{64} = 4$.
Key idea
$\sqrt[3]{n}$ is the number whose cube is $n$. The perfect cubes are $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, \ldots$
Unlike square roots, cube roots of negative numbers exist. Three negative factors make a negative, so $(-4)^{3} = -64$ and $\sqrt[3]{-64} = -4$.
Worked examples
Worked example: A perfect cube
Find $\sqrt[3]{125}$.
- Which number, used three times as a factor, gives 125?
- $5 \cdot 5 \cdot 5 = 125$.
Answer: $5$
Worked example: A negative number
Find $\sqrt[3]{-27}$.
- $3^{3} = 27$, so try $-3$.
- $(-3)(-3)(-3) = 9 \cdot (-3) = -27$.
Answer: $-3$
Worked example: A fraction
Find $\sqrt[3]{\dfrac{8}{27}}$.
- Take the cube root of the top and of the bottom: $\sqrt[3]{8} = 2$ and $\sqrt[3]{27} = 3$.
- Check: $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{8}{27}$.
Answer: $\frac{2}{3}$
Most numbers are not perfect cubes. Their cube roots still have a place between whole numbers: $3^{3} = 27$ and $4^{3} = 64$, so $\sqrt[3]{50}$ is between 3 and 4.
Common mistake
Dividing by 3, as in $\sqrt[3]{27} = 9$. Check by cubing: $9^{3} = 729$. The cube root of 27 is 3, because $3 \cdot 3 \cdot 3 = 27$.