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**Hint:**We are given a term as ${{4}^{\dfrac{4}{3}}}$, we have to change it into radical form. We will learn what form is the number given to us in and what are radical, then we will use that ${{({{x}^{a}})}^{b}}={{x}^{ab}}$. We will learn about exponential function to finite ${{4}^{\dfrac{4}{3}}}$into radical form we will split its power and make copies as much as the denominator of the power fraction, then we will simplify.

**Complete step-by-step solution:**

We are given ${{4}^{\dfrac{4}{3}}}$. We can see that we have a term and it is raised to some power just like ${{a}^{b}}$, which is just an exponential term. So, we are given an exponential term. Whose base is $4$ and power is \[\dfrac{4}{3}\]

We are asked to change it into the radical form. The radical form is a form that contains a square root. Radical in a number or expression that stays inside the radical symbolise for different power we write it differently. If we have square root $({{x}^{\dfrac{1}{2}}})$we use just $\sqrt{x}$.

If we have cube root $({{x}^{\dfrac{1}{3}}})$ we use $\sqrt[3]{x}$

If we have $({{x}^{\dfrac{1}{4}}})$ we use $\sqrt[4]{x}$

Now we will work on our problem. We have ${{4}^{\dfrac{4}{3}}}$.

We can see that \[\dfrac{4}{3}\] can be written as $4\times \dfrac{1}{3}$

So, ${{\left( {{4}^{4}} \right)}^{\dfrac{4}{3}}}$

Now as we know that ${{({{n}^{a}})}^{b}}={{x}^{ab}}$ or \[{{x}^{ab}}={{({{y}^{a}})}^{b}}\]

So we will use this on ${{4}^{\dfrac{4}{3}}}$

So, ${{4}^{4\times \dfrac{4}{3}}}$= ${{({{4}^{4}})}^{{}^{1}/{}_{3}}}$

Now we will use the rule we discuss above to change the expression into radical.

We know $({{x}^{\dfrac{1}{3}}})$= $\sqrt[3]{x}$

So as we have ${{4}^{4\times \dfrac{4}{3}}}$ so using above formula we get

${{({{4}^{4}})}^{{}^{1}/{}_{3}}}$$=\sqrt[3]{{{4}^{4}}}$

now we get we have to find the cube root of ${{4}^{4}}$

so will expand ${{4}^{4}}$ into factors.

So ${{4}^{4}}=4\times 4\times 4\times 4=256$

Hence, we get

${{({{4}^{4}})}^{{}^{1}/{}_{3}}}$$=\sqrt[3]{{{4}^{4}}}$

$\sqrt[3]{256}$

Now we simplify it

we take $3$ pair of same term out with cube root

as ${{4}^{4}}=4\times 4\times 4\times 4$

**So, $3$ will taken out and become**

${{({{4}^{4}})}^{{}^{1}/{}_{3}}}$$=4\sqrt[3]{4}$

${{({{4}^{4}})}^{{}^{1}/{}_{3}}}$$=4\sqrt[3]{4}$

**Note:**It is necessary to split term into denominator and numerator for carrying working exponential function as defined as ${{x}^{4}}$these are properties as exponential.

$\begin{align}

& {{x}^{a}}\times {{x}^{b}}={{x}^{a+b}} \\

& \dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}} \\

& {{({{x}^{a}})}^{b}}={{x}^{ab}} \\

\end{align}$

These also work when we use them to find the radical, we can simplify one term using these before changing them to the radical form.

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