How do you simplify ${{\left( 4{{a}^{2b}} \right)}^{4}}{{\left( \dfrac{-{{a}^{3}}}{2b} \right)}^{2}}$

Hint: Now to simplify the given expression we will first use the property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ and open the bracket. Now we will combine the same variables by using the multiplication and division properties of indices. Hence we get the required expression in simplified form.

Complete step by step solution:
We will first understand the theory of exponents and various rules related to it.
Exponents are nothing but numbers or variables raised to something.
Now consider an exponent ${{a}^{3}}$ here we have 3 raised to a.
Now the number which is raised is called the power and it denotes the number of times the term is multiplied by itself.
Hence we can expand ${{a}^{3}}=a\times a\times a$ .
Now we can perform and simplify the exponents using different rules.
Now let us first understand the multiplication rule. Suppose we have two exponents in multiplication with the same base term. Then the multiplication is given as, ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$
Now let us understand the division rule. Suppose we have two exponents in division with the same base term. Then the division is given as, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ . Now suppose we have an exponent raised to another number. Then the value of such exponent is given by property ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ .
Now also note that using division rule and multiplication rule we have ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ .
Now consider the given expression ${{\left( 4{{a}^{2b}} \right)}^{4}}{{\left( \dfrac{-{{a}^{3}}}{2b} \right)}^{2}}$
Using the exponent law we get,
$\begin{align}
  & \Rightarrow 4{{a}^{2b\times 8}}\left( \dfrac{-{{a}^{3\times 2}}}{2{{b}^{2}}} \right) \\
 & \Rightarrow 4{{a}^{8b}}\times \dfrac{-{{a}^{6}}}{2{{b}^{2}}} \\
\end{align}$
Now using the multiplication rule we get,
$\Rightarrow \dfrac{-4{{a}^{8b-6}}}{2{{b}^{2}}}$
Hence on simplifying we get,
$\Rightarrow \dfrac{2{{a}^{8b+6}}}{{{b}^{2}}}$
Hence the given expression can be written as $\dfrac{2{{a}^{8b+6}}}{{{b}^{2}}}$

Note:
Now note that the power of a number can be zero negative or positive number. Let us take an example of each and understand the meaning . ${{2}^{4}}=2\times 2\times 2\times 2=16$ , ${{2}^{-4}}=\dfrac{1}{{{2}^{4}}}=\dfrac{1}{16}$ and for any number with 0 in its power has value equal to 1.