How do you simplify $b{{a}^{4}}.{{\left( 2b{{a}^{4}} \right)}^{-3}}$ and write it using only positive exponents?
 (a) Using algebraic identities
(b) Using properties of exponents
(c) Using trigonometric quantities
(d) a and b both

Hint: To work with the problem, we will use the property of exponents. To start with, we will try to work with the part of the term where the negative power is given. Once we do that, we can multiply that with the term left and get the needed solution using the properties of the exponents.

Complete step by step answer:
According to the question, we are trying to simplify $b{{a}^{4}}.{{\left( 2b{{a}^{4}} \right)}^{-3}}$.
So, to start with, let us go on with the second part of the term.
It is given as, ${{\left( 2b{{a}^{4}} \right)}^{-3}}$,
Now, we will be using one property of exponents said to be, ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$.
This is called the negative power rule.
Writing ${{\left( 2b{{a}^{4}} \right)}^{-3}}$in that form, $\dfrac{1}{{{\left( 2b{{a}^{4}} \right)}^{3}}}$ .
Now, by simplifying the denominator gives us, $\dfrac{1}{8{{b}^{3}}{{a}^{12}}}$.
In this we have applied the power rule as, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .
Again, $b{{a}^{4}}.{{\left( 2b{{a}^{4}} \right)}^{-3}}$can be now written as, $\dfrac{b{{a}^{4}}}{{{\left( 2b{{a}^{4}} \right)}^{3}}}=\dfrac{b{{a}^{4}}}{8{{b}^{3}}{{a}^{12}}}$.
Now, applying quotient rule, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ ,
Let us write it using the property.
$\dfrac{b{{a}^{4}}}{8{{b}^{3}}{{a}^{12}}}=\dfrac{{{b}^{1-3}}{{a}^{4-12}}}{8}$
After more simplification, we get, $\dfrac{{{b}^{-2}}{{a}^{-8}}}{8}$.
Again, applying the negative power rule, ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$, we are getting,
$\dfrac{{{b}^{-2}}{{a}^{-8}}}{8}=\dfrac{1}{8{{b}^{2}}{{a}^{8}}}$.

Hence, the solution is, (d) a and b both.

Note:
When we combine like terms by adding and subtracting, we need to have the same base with the same exponent. But when you multiply and divide, the exponents may be different, and sometimes the bases may be different. In every case the bases of the terms also should be just the same.