Question

What is the cube root of -216?

Answer 1

Hint: Cube root of a number is a value which when multiplied by itself thrice or three times produces the original value. To find the cube root of -216, that is, $\sqrt[3]{-216}$ , we have to split this as $\sqrt[3]{\left( -1 \right)\times 216}$ . By using $\sqrt[n]{ab}=\sqrt[n]{a}\times \sqrt[n]{b}$ , we can write $\sqrt[3]{\left( -1 \right)\times 216}=\sqrt[3]{-1}\times \sqrt[3]{216}$ . We have to find the cube root of -1 and 216 separately and substitute it in this form.

Complete step by step solution:
We have to find the cube root of -216. We know that the cube root of a number is a value which when multiplied by itself thrice or three times produces the original value. Cube root is denoted as $\sqrt[3]{{}}$ . Let us consider the cube root of${{m}^{3}}$ , that is, $\sqrt[3]{{{m}^{3}}}$ . We know that $m\times m\times m={{m}^{3}}$ . Hence, the cube root of ${{m}^{3}}$ will be m.
Now, let us consider $\sqrt[3]{-216}$ . We can write this as $\sqrt[3]{\left( -1 \right)\times 216}$ .
We know that $\sqrt[n]{ab}=\sqrt[n]{a}\times \sqrt[n]{b}$ . Hence, we can write $\sqrt[3]{\left( -1 \right)\times 216}$ as
$\sqrt[3]{\left( -1 \right)\times 216}=\sqrt[3]{-1}\times \sqrt[3]{216}...\left( i \right)$
Let us first find the cube root of -1. We know that $-1\times -1\times -1=-1$ . Therefore, we can write
$\sqrt[3]{-1}=-1...\left( ii \right)$
Now, let us find the cube root of 216. For this, let us find the prime factors of 216.
$\begin{align}
  & 2\left| \!{\underline {\,
  216 \,}} \right. \\
 & 2\left| \!{\underline {\,
  108 \,}} \right. \\
 & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }1 \\
\end{align}$
We can write 216 as $2\times 2\times 2\times 3\times 3\times 3$ . We have to combine each triple.
$216=\boxed{2\times 2\times 2}\times \boxed{3\times 3\times 3}$
Hence, we can write $\sqrt[3]{216}=2\times 3=6...\left( iii \right)$ .
Let us substitute (ii) and (iii) in (i).
$\Rightarrow \sqrt[3]{-216}=-1\times 6=-6$
Hence, the cube root of -216 is -6.

Note: Students must know the properties of roots and how to find the prime factors of 216. We can also denote cube root of a number, say m, as m to the power of $\dfrac{1}{3}$ or ${{m}^{\dfrac{1}{3}}}$ .
Therefore, we can write $\sqrt[3]{-216}={{\left( -216 \right)}^{\dfrac{1}{3}}}$.