Question

The square root of $64$ divided by the cube root of $64$ is:
(i) $64$
(ii) $2$
(iii) $\dfrac{1}{2}$

Answer 1

Hint: We know that the number $64$ is the square of $8$ and it is also the cube of $4$. Use these results to simplify the operation given in the question and find the answer.

Complete step-by-step solution:
According to the question, we have to determine the result when the square root of $64$ is divided by the cube root of $64$.
So we have to determine the value of $\dfrac{{\sqrt {64} }}{{\sqrt[3]{{64}}}}$.
We know that the number $64$ is the square of $8$ and it is also the cube of $4$ as shown below:
$ \Rightarrow 64 = {8^2}$ and $64 = {4^3}$
Taking square root and cube root on both the sides respectively, we’ll get:
$ \Rightarrow \sqrt {64} = 8$ and $\sqrt[3]{{64}} = 4$
Using these results to determine the value of $\dfrac{{\sqrt {64} }}{{\sqrt[3]{{64}}}}$, we’ll get:
$
   \Rightarrow \dfrac{{\sqrt {64} }}{{\sqrt[3]{{64}}}} = \dfrac{8}{4} \\
   \Rightarrow \dfrac{{\sqrt {64} }}{{\sqrt[3]{{64}}}} = 2
 $
Therefore the square root of $64$ divided by the cube root of $64$ is 2.

(ii) is the correct option.

Note: Square root and cube root of a number can be either rational or irrational numbers. For example,
$ \Rightarrow \sqrt {16} = 4,{\text{ }}\sqrt {100} = 10,{\text{ }}\sqrt[3]{{216}} = 6,{\text{ }}\sqrt[3]{{15.625}} = 2.5$
If a number is not a perfect square and we take its square root, then the result will always be an irrational number. For example $\sqrt 2 ,{\text{ }}\sqrt {10} ,{\text{ }}\sqrt {17} $ are all irrational numbers.
Similarly if a number is not a perfect cube and we take its cube root, then also the result will always be an irrational number. For example $\sqrt[3]{7},{\text{ }}\sqrt[3]{{15}},{\text{ }}\sqrt[3]{{21}}$ are all irrational numbers.
Square roots and cube roots of prime numbers are always irrational.
Taking the square root of a number is equivalent of raising it to power $\dfrac{1}{2}$, taking the cube root is similar to raising power $\dfrac{1}{3}$ and taking the 4th root is similar to raising to raising power $\dfrac{1}{4}$. Likewise, taking the nth root is similar to raising power $\dfrac{1}{n}$ and so on.
$ \Rightarrow \sqrt {13} = {13^{\dfrac{1}{2}}},{\text{ }}\sqrt[3]{8} = {8^{\dfrac{1}{3}}},{\text{ }}\sqrt[4]{{20}} = {20^{\dfrac{1}{4}}},{\text{ }}\sqrt[n]{{46}} = {46^{\dfrac{1}{n}}}$