How do you simplify the given term: $\dfrac{x-3}{2x-8}\cdot \dfrac{6{{x}^{2}}-96}{{{x}^{2}}-9}$?

Hint: We start solving the problem by making the necessary arrangements in the given term and then making use of the fact that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ to proceed through the problem. We then cancel the terms that were present common in both numerator and denominator to proceed further through the problem. We then make the necessary arrangements in the numerator and then make use of the that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$ which gives the required simplified form after cancelling the common terms in it.

Complete step by step answer:
According to the problem, we are asked to simplify the given term $\dfrac{x-3}{2x-8}\cdot \dfrac{6{{x}^{2}}-96}{{{x}^{2}}-9}$.
Let us assume $d=\dfrac{x-3}{2x-8}\cdot \dfrac{6{{x}^{2}}-96}{{{x}^{2}}-9}$.
\[\Rightarrow d=\dfrac{x-3}{2\left( x-4 \right)}\cdot \dfrac{6\left( {{x}^{2}}-16 \right)}{{{x}^{2}}-9}\] ---(1).
We know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Let us use this result in equation (1).
\[\Rightarrow d=\dfrac{x-3}{2\left( x-4 \right)}\cdot \dfrac{6\left( x-4 \right)\left( x+4 \right)}{\left( x-3 \right)\left( x+3 \right)}\] ---(2).
Now, let us cancel the terms that were present common in both numerator and denominator of equation (2).
\[\Rightarrow d=\dfrac{3\left( x+4 \right)}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{3x+12}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{3x+9+3}{\left( x+3 \right)}\] ---(3).
We know that $\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}$. Let us use this result in equation (3).
\[\Rightarrow d=\dfrac{3x+9}{\left( x+3 \right)}+\dfrac{3}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{3\left( x+3 \right)}{\left( x+3 \right)}+\dfrac{3}{\left( x+3 \right)}\] ---(4).
Now, let us cancel the common terms present in equation (4).
\[\Rightarrow d=3+\dfrac{3}{\left( x+3 \right)}\].
So, we have found the simplified form of the given term $\dfrac{x-3}{2x-8}\cdot \dfrac{6{{x}^{2}}-96}{{{x}^{2}}-9}$ as \[3+\dfrac{3}{\left( x+3 \right)}\].

$\therefore $ The simplified form of the given term $\dfrac{x-3}{2x-8}\cdot \dfrac{6{{x}^{2}}-96}{{{x}^{2}}-9}$ is \[3+\dfrac{3}{\left( x+3 \right)}\].

Note: We should perform each step carefully in order to avoid calculation mistakes and confusion while solving this problem. We can also simplify the given term after equation (2) as shown below:
\[\Rightarrow d=\dfrac{3\left( x+4 \right)}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{3x+12}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{4x+12-x}{\left( x+3 \right)}\] ---(5).
We know that $\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}$. Let us use this result in equation (5).
\[\Rightarrow d=\dfrac{4x+12}{\left( x+3 \right)}-\dfrac{x}{\left( x+3 \right)}\].
\[\Rightarrow d=\dfrac{4\left( x+3 \right)}{\left( x+3 \right)}-\dfrac{x}{\left( x+3 \right)}\] ---(6).
Now, let us cancel the common terms present in equation (6).
\[\Rightarrow d=4-\dfrac{x}{\left( x+3 \right)}\].