Question

How do you simplify ${64^{\dfrac{2}{3}}}$.

Answer 1

Hint: This sum is basically a combination of $2$ concepts. The student should know the squares and cubes in order to simplify the sum. Further the student has to use the property of Indices i.e. ${({a^x})^y} = {a^{x \times y}}$. With this the student can easily simplify these types of sums and several other sums that involve similar methodology.

Complete step-by-step answer:
The first step in solving such types of problems is to simplify the sum as much as possible. From the given sum , we can say that $64$ is a square of $8$.
We can simplify the given numerical and write it in terms of $8$.
$\Rightarrow {({8^2})^{\dfrac{2}{3}}}..............(1)$
Now to simplify further, we will be using the properties of indices i.e. ${({a^x})^y} = {a^{x \times y}}$. The next step is as follows
$\Rightarrow {(8)^{\dfrac{{2 \times 2}}{3}}} = {(8)^{\dfrac{4}{3}}}..............(2)$
But we can also say that $8$ is a cube of $2$, or cube root of $8$ is $2$.
$8$ Can be further written as $8 = {2^3}...........(3)$
Substituting the value of $8$ from equation $3$ in equation $2$ we get the final answer.
$\Rightarrow {(8)^{\dfrac{4}{3}}} = {2^{3 \times \dfrac{4}{3}}}..............(4)$
Thus the final answer for the given numerical is ${2^4} = 16$.

Note: These sums look easy to solve as it has been solved by implementing a proper methodology. On the contrary if the student tries to calculate and then put the values it would have become difficult to solve. Thus it is advisable that students make use of indices to solve such sums. Also the students should be thorough with squares, cubes, square root and cube root. If these are on the tips the sum would become extremely easy to solve. It may so happen that the students could solve the sum in their mind within a matter of a few seconds.