Answer
$\color{blue}{(-\infty, -1) \cup (-1, 6) \cup (6, +\infty)}$
Work Step by Step
Factor the denominator to obtain:
$=\dfrac{3}{(x-6)(x+1)}$
The denominator of a rational expression is not allowed to be equal to zero as it will make the expression undefined.
Find the values of $x$ for which the denominator is equal to zero by equating the denominator to zero:
$(x-6)(x+1)= 0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
\begin{array}{ccc}
&x-6 = 0 &\text{ or } &x+1=0
\\&x=6 &\text{or} &x=-1
\end{array}
This means that the value of $x$ can be any real number except $6$ and $-1$.
Therefore, the domain of the given rational expression is the set of real numbers except $6$ and $-1$.
In interval notation, the domain is:
$\color{blue}{(-\infty, -1) \cup (-1, 6) \cup (6, +\infty)}$
