Question

How do you simplify \[{216^{\dfrac{2}{3}}}\]?

Answer 1

Hint: To simplify \[{216^{\dfrac{2}{3}}}\]
In this type of question, we should try to factorize the inner number. If there are any common factors then we can write in terms of power. In this question, we will try to use one of the properties of simple algebra that is
\[a \times a \times a \times a............n{\rm{ }}times = {\left( a \right)^n}\]
By using these properties we should break \[216\] and by using the value of n we should simplify this expression to get a rational number. In prime factorization we try to break the given number into prime factor means factors which are only divisible by itself and 1.

Complete step by step solution:
Firstly, we will try to break \[216\] into simple prime factors. Like,
\[216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\]
Then, by using basic property that is \[a \times a \times a \times a............n{\rm{ }}times = {\left( a \right)^n}\]
We can write it as
\[216 = {\left( 2 \right)^3}{\left( 3 \right)^3}\]
Now, we have to find the value of \[{216^{\dfrac{2}{3}}}\]
\[ \Rightarrow {\left( {216} \right)^{\dfrac{2}{3}}} = {\left( {{{\left( 2 \right)}^3}{{\left( 3 \right)}^3}} \right)^{\dfrac{2}{3}}}\]
Now, we will use one other basic property that is \[{\left( {{a^b}} \right)^c} = {a^{b \times c}}\]
By applying above property we get
\[\begin{array}{l}
{\left( {216} \right)^{\dfrac{2}{3}}} = \left( {{2^{3 \times \dfrac{2}{3}}}} \right) \times \left( {{3^{3 \times \dfrac{2}{3}}}} \right)\\
 = {2^2} \times {3^2}\\
 = 4 \times 9\\
 = 36
\end{array}\]

So, we get our answer as \[36\]

Note:
We must take care that while doing factorization each prime factor should be written, we shouldn’t skip any prime factor. Prime factorization of any irrational no. is not possible. So, the number must be rational so that by using the above properties we can simplify it. While writing prime factors don’t consider 1 as a factor because 1 is not a prime factor.