Simplify-

(a) $\sqrt{75a^9}$

(b)$\sqrt[3]{128m^{11}}$

(c)$\sqrt[4]{62n^7}$

(a) We will write the radical in form of the product of two radicals with one being the largest perfect square factor of the number. Then we will simplify it to get the value.

(b) We will write the radical in form of the product of two radicals with one being the largest perfect cube factor of the number. Then we will simplify it to get the value.

(c) We will write the radical in form of the product of two radicals with one being the largest perfect fourth power factor of the number. Then we will simplify it to get the value.

We have $\sqrt{75a^9}$.

We have $25a^8$as the greatest perfect square factor of $75a^9$.

So this can be written as $\sqrt{25a^8\times 3a}$.

Using the product rule, we get

$$\begin{gathered} =\sqrt{25a^8}\times \sqrt{3a} \\ =\sqrt{{\left(5a^4\right)}^2}\times \sqrt{3a} \\ =5a^4\sqrt{3a} \end{gathered}$$

Thus we get the value as$5a^4\sqrt{3a}$.

We have $\sqrt[3]{128m^{11}}$.

We have $64m^9$as the greatest perfect cube factor of $128m^{11}$.

So this can be written as $\sqrt[3]{64m^9\times 2m^2}$.

Using the product rule, we get

$$\begin{gathered} =\sqrt[3]{64m^9}\times \sqrt[3]{2m^2} \\ =\sqrt[3]{{\left(4m^3\right)}^3}\times \sqrt[3]{2m^2} \\ =4m^3\sqrt[3]{2m^2} \end{gathered}$$

Thus we get the value as$4m^3\sqrt[3]{2m^2}$.

We have $\sqrt[4]{62n^7}$.

We have $n^4$as the greatest perfect fourth power factor of $62n^7$.

So this can be written as $\sqrt[4]{n^4\times 62n^3}$.

Using the product rule, we get

$$\begin{gathered} =\sqrt[4]{n^4}\times \sqrt[4]{62n^3} \\ =\sqrt[4]{{\left(n\right)}^4}\times \sqrt[4]{62n^3} \\ =\left|n\right|\sqrt[4]{62n^3} \end{gathered}$$

Thus we get the value as$\left|n\right|\sqrt[4]{62n^3}$.