Question

How do you find the cube roots $\sqrt[3]{27}$?

Answer 1

Hint: We try to form the indices formula for the value 3. This is a cube root of 432. We find the prime factorisation of 27. Then we take one digit out of the three same number of primes. There will be no odd number of primes remaining in the root which can’t be taken out. We keep the cube root in its simplest form.

Complete step by step answer:
We need to find the value of the algebraic form of $\sqrt[3]{27}$. This is a cube root form.
The given value is the form of indices. We are trying to find the root value of 27.
We know the theorem of indices \[{{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}\]. Putting value 3 we get \[{{a}^{\dfrac{1}{3}}}=\sqrt[3]{a}\].
We need to find the prime factorisation of the given number 27.
$\begin{align}
  & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
Therefore, \[27=3\times 3\times 3\].
For finding the cube root, we need to take one digit out of the three same number of primes.
This means in the cube root value of \[27=3\times 3\times 3\], we will take out one 3 from the multiplication.
So, $\sqrt[3]{27}=\sqrt[3]{3\times 3\times 3}=3$. Basically 27 is the cube of 3.
We can also use the theorem of indices \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\]. We know that $27={{\left( 3 \right)}^{3}}$.
We need to find $\sqrt[3]{27}$ which gives $27={{\left( 3 \right)}^{3}}={{\left[ {{3}^{3}} \right]}^{\dfrac{1}{3}}}={{\left( 3 \right)}^{3\times \dfrac{1}{3}}}=3$.
Therefore, the value of $\sqrt[3]{27}$ is 3.

Note:
We can also use the variable form where we can take $x=\sqrt[3]{27}$. But we need to remember that we can’t use the cube on both sides of the equation $x=\sqrt[3]{27}$ as in that case we are taking two extra values as a root value. Then this linear equation becomes a cubic equation.