How do you simplify the expression \[\dfrac{12{{a}^{-3}}{{b}^{9}}}{21{{a}^{2}}{{b}^{-5}}}\] using the properties?

Hint: First of all write the given numbers 12 and 21, in the numerator and denominator, respectively as the product of their prime factors. Now, cancel the prime factors that are common and simplify the expression. Now, use the formulas: - \[\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}\] to simplify the exponential terms and get the answer.

Complete step by step solution:
Here, we have been provided with the expression \[\dfrac{12{{a}^{-3}}{{b}^{9}}}{21{{a}^{2}}{{b}^{-5}}}\] and we are asked to simplify it using the properties and formulas of exponents. Let us assume the given expression as ‘E’, so we have,
\[\Rightarrow E=\dfrac{12{{a}^{-3}}{{b}^{9}}}{21{{a}^{2}}{{b}^{-5}}}\]
Now, we can write this expression as: -
\[\Rightarrow E=\dfrac{12}{21}\times \dfrac{{{a}^{-3}}}{{{a}^{2}}}\times \dfrac{{{b}^{9}}}{{{b}^{-5}}}\]
Let us simplify the ratio \[\dfrac{12}{21}\] first. To simplify this ratio we need to cancel the common factors present in them. So, let us write these numbers as the product of their prime factors.
\[\Rightarrow \dfrac{12}{21}=\dfrac{3\times 2\times 2}{3\times 7}\]
Clearly, we can see that the only common factor is 3, so cancelling this common factor, we get,
\[\begin{align}
  & \Rightarrow \dfrac{12}{21}=\dfrac{2\times 2}{7} \\
 & \Rightarrow \dfrac{12}{21}=\dfrac{4}{7} \\
\end{align}\]
Substituting this value in the expression ‘E’, we get,
\[\Rightarrow E=\dfrac{4}{7}\times \dfrac{{{a}^{-3}}}{{{a}^{2}}}\times \dfrac{{{b}^{9}}}{{{b}^{-5}}}\]
Using the basic formulas of exponents given as: - \[\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}\], we have,
\[\begin{align}
  & \Rightarrow E=\dfrac{4}{7}\times {{a}^{-3-2}}\times {{b}^{9-\left( -5 \right)}} \\
 & \Rightarrow E=\dfrac{4}{7}\times {{a}^{-5}}\times {{b}^{9+5}} \\
 & \Rightarrow E=\dfrac{4}{7}\times {{a}^{-5}}\times {{b}^{14}} \\
 & \Rightarrow E=\dfrac{4{{a}^{-5}}{{b}^{14}}}{7} \\
\end{align}\]
Using the formula: - \[{{x}^{-m}}=\dfrac{1}{{{x}^{m}}}\], we can further simplify the expression as: -
\[\Rightarrow E=\dfrac{4{{b}^{14}}}{7{{a}^{5}}}\]
Hence, the above expression is our answer.

Note: One may note that the expression \[\dfrac{4{{b}^{14}}}{7{{a}^{5}}}\] cannot be simplified further because there are no common factors present that can be cancelled. Here, we have used some basic formulas of the topic ‘exponents and powers’ which must be remembered to solve the question. You must remember formulas like: - \[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\], \[{{x}^{m}}\div {{x}^{n}}={{x}^{m-n}}\], \[{{\left( {{x}^{m}} \right)}^{n}}={{x}^{m\times n}}\], \[{{x}^{-m}}=\dfrac{1}{{{x}^{m}}}\] etc, as they are used in certain other topics of mathematics. Remember the method of prime factorization and how to write a composite number as the product of its prime factors.