# What are the cube roots of 16?

Apr 4, 2017

$x = \sqrt[3]{16}$ or $\frac{- \sqrt[3]{16} + i \sqrt[3]{16} \sqrt{3}}{2}$ or $\frac{- \sqrt[3]{16} - i \sqrt[3]{16} \sqrt{3}}{2}$

#### Explanation:

${x}^{3} = 16 \implies {x}^{3} - 16 = 0$ i.e.

${x}^{3} - {\left(\sqrt[3]{16}\right)}^{3} = 0$, now let $\sqrt[3]{16} = k$, then we have

${x}^{3} - {k}^{3} = 0$

or $\left(x - k\right) \left({x}^{2} + k x + {k}^{2}\right) = 0$

Hence $x - k = 0$ i.e. $x = k$ i.e. $x = \sqrt[3]{16}$

or ${x}^{2} + k x + {k}^{2} = 0$ i.e. $x = \frac{- k \pm \sqrt{{k}^{2} - 4 {k}^{2}}}{2}$

i.e. $x = \frac{- k \pm i k \sqrt{3}}{2} = k \left(\frac{- 1 \pm i \sqrt{3}}{2}\right)$

i.e. $x = \sqrt[3]{16} \left(\frac{- 1 \pm i \sqrt{3}}{2}\right)$

i.e. $x = \frac{- \sqrt[3]{16} + i \sqrt[3]{16} \sqrt{3}}{2}$ or $x = \frac{- \sqrt[3]{16} - i \sqrt[3]{16} \sqrt{3}}{2}$

Note - Observe that if $1 , \omega , {\omega}^{2}$ are cube roots of $1$,

cube roots of $16$ are $k , k \omega , k {\omega}^{2}$, where $k = \sqrt[3]{16}$