Simplify-

(a) $\sqrt{b^5}$

(b) $\sqrt[4]{y^6}$

(c)$\sqrt[3]{z^5}$

(a) We will write the radical in form of 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 here, write the root of the number as the power of the number. Then after simplifying the expression, we will get the value.

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

We have $\sqrt{b^5}$.

We have $b^4$as the largest perfect square factor of $b^5$.

So this can be written as $\sqrt{b^4\times b}$.

We know that ${\left(b^2\right)}^2=b^4$.

Using the product rule we get,

$\sqrt{b^4}\times \sqrt{b}$

Now using this value, we can write the expression as$b^2\sqrt{b}$.

We have $\sqrt[4]{y^6}$.

We know that fourth root of a number is equal to the number raised to one fourth power.

So using this, we can write the expression as ${\left(y\right)}^{6\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$.

Simplifying it will give us ${\left(y\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

localid="1645704009057" role="math" $$\begin{gathered} {\left(y\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(y\right)}^{1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ ={\left(y\right)}^1\times {\left(y\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =y\times \sqrt{y} \\ =y\sqrt{y} \end{gathered}$$

We have $\sqrt[3]{z^5}$.

We have $z^3$as the largest perfect cube factor of $z^5$.

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

Using product rule, we get $\sqrt[3]{z^3}\times \sqrt[3]{z^2}$.

Simplifying this will give us$z\sqrt[3]{z^2}$.