Question

How do you write ${64^{\dfrac{5}{6}}}$ in radical form?

Answer 1

Hint: We will start by factorising the term inside the square root. Then we will separate all the like terms and all the alike terms. Then we will check if any squares are present, if present then we will take those terms out of the square root.

Complete step-by-step solution:
First we will start off by writing the term inside the square root to simplify the terms. Then next we will start by forming factors inside the root to further simplify.
$
   \Rightarrow{64^{\dfrac{5}{6}}} \\
   \Rightarrow{\left( {\sqrt[6]{{64}}} \right)^5} \\
 $
Now we separate the like terms.
$ \Rightarrow{\left( {\sqrt {(32 \times 2)} } \right)^5}$
Now we check if any square root is present to take them out of the square root.
$
   \Rightarrow{\left( {\sqrt[6]{{(16 \times 2 \times 2)}}} \right)^5} \\
   \Rightarrow{\left( {\sqrt[6]{{(4 \times 4 \times 2 \times 2)}}} \right)^5} \\
   \Rightarrow{\left( {\sqrt[6]{{(2 \times 2 \times 2 \times 2 \times 2 \times 2)}}} \right)^5} \\
   \Rightarrow{\left( {{2^6}} \right)^{\dfrac{5}{6}}} \\
   \Rightarrow{2^5} \\
   \Rightarrow32 \\
 $
So, ${32^{\dfrac{4}{5}}}$ in simplified form is written as $16$.

Additional information: Expressing in simplest radical form just means simplifying a radical so that there are no more square roots, cube roots, fourth roots, etc left to find. It also means removing any radicals in the denominator of fraction. Basically, finding the ${n^{th}}$ root of a positive number is the opposite of raising the number to the power $n$, so they effectively cancel each other out. These expressions have the same value. The square root of the product is the same as the product of the square roots. You can use the following property to simplify a square root: $\sqrt {ab} \Rightarrow\sqrt a \times \sqrt b $.

Note: While factorising the terms, factorise until you cannot factorise any further. Also, while factorising takes into consideration the signs of the terms. While searching for square roots, searching for even powers, it makes the process easier.