Simplify and express the answer in positive exponent form $\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}}$

Hint: In this question, we are given a fraction in which nearly all the terms are raised to some power. The power of a number is defined as the number of times the base number is multiplied by itself. We have to simplify this fraction and express it in the form of a positive exponent, so instead of simply multiplying the terms and then canceling out the factors, we will express all the terms as a product of their prime factors and then write them in exponential form. Then using the laws of exponents, we can solve the obtained expression.

Complete step-by-step solution:
We have to simplify $\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}}$ so we will write all the terms as the product of their prime factors.
On the prime factorization of 6, 12 and 36, we see that –
$
  6 = 2 \times 3 \\
  12 = 2 \times 2 \times 3 \\
  36 = 2 \times 2 \times 3 \times 3 \\
 $
So, we can write $\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}}$ as $\dfrac{{2 \times 2 \times 3 \times 3 \times {{( - 1)}^2}{{(2 \times 3)}^2} \times {3^6}}}{{{{(2 \times 2 \times 3)}^3} \times {3^5}}}$
It can be rearranged as –
$ \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{{2^2} \times {2^2} \times {3^2} \times {3^2} \times {3^6}}}{{{{({2^2})}^3} \times {3^3} \times {3^5}}}$
We know that \[{(ab)^m} = {a^m}{b^m}\]
$ \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{{2^2} \times {3^2} \times {2^2} \times {3^2} \times {3^6}}}{{{{({2^2})}^3} \times {3^3} \times {3^5}}}$
We also know that ${({a^n})^m} = {a^{n \times m}}$ and ${a^n} \times {a^m} = {a^{m + n}}$
$
   \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{{2^{2 + 2}} \times {3^{2 + 2 + 6}}}}{{{2^{2 \times 3}} \times {3^{3 + 5}}}} \\
   \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{{2^4} \times {3^{10}}}}{{{2^6} \times {3^8}}} \\
 $
As we know that $\dfrac{{{a^n}}}{{{a^m}}} = {a^{n - m}}$ , we get –
$
  \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = ({2^{4 - 6}})({3^{10 - 8}}) \\
   \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = {2^{ - 2}}{3^2} \\
   \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = {(\dfrac{3}{2})^2}\,\,\,\,\,({a^{ - n}} = \dfrac{1}{{{a^n}}}) \\
 $
As the answer obtained has a power +2, so it is a positive exponent.

Hence $\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}}$ is expressed in positive exponent form as ${(\dfrac{3}{2})^2}$ .

Note: Various laws of exponents were used in solving the given question so the process turned out to be quite long. So, instead of solving the question this way, we can solve it by simplifying the exponent terms and then multiplying all the terms in the numerator with each other and those in the denominator with each other. Then we will cancel out the common factors of the numerator and the denominator and express the obtained fraction in exponent form. It can be done as –
$\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{36 \times 36 \times {3^6}}}{{12 \times 12 \times 12 \times {3^5}}}$
Now, 36 can be written as the product of 3 and 12 –
$\dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{3 \times 12 \times 3 \times 12 \times {3^6}}}{{12 \times 12 \times 12 \times {3^5}}}$
To simplify the above expression, we cancel out the common factors. 12 can be written as the product of 3 and 4 and on simplifying ${3^6}$ and ${3^5}$ , we get –
$ \Rightarrow \dfrac{{36 \times {{( - 6)}^2} \times {3^6}}}{{{{12}^3} \times {3^5}}} = \dfrac{{3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}}{{3 \times 4 \times 3 \times 3 \times 3 \times 3 \times 3}}$