Find the minimum value of
\[\frac{\left( x + \dfrac{1}{x} \right)^6 - \left( x^6 + \dfrac{1}{x^6} \right) - 2}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)}\]for $x > 0.$
Let
\[f(x) = \frac{\left( x + \dfrac{1}{x} \right)^6 - \left( x^6 + \dfrac{1}{x^6} \right) - 2}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)}.\]We can write
\[f(x) = \frac{\left( x + \dfrac{1}{x} \right)^6 - \left( x^6 + 2 + \dfrac{1}{x^6} \right)}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)} = \frac{\left( x + \dfrac{1}{x} \right)^6 - \left( x^3 + \dfrac{1}{x^3} \right)^2}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)}.\]By difference of squares,
\begin{align*}
f(x) &= \frac{\left[ \left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right) \right] \left[ \left( x + \dfrac{1}{x} \right)^3 - \left( x^3 + \dfrac{1}{x^3} \right) \right]}{\left( x + \dfrac{1}{x} \right)^3 + \left( x^3 + \dfrac{1}{x^3} \right)} \\
&= \left( x + \dfrac{1}{x} \right)^3 - \left( x^3 + \dfrac{1}{x^3} \right) \\
&= x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} - x^3 - \frac{1}{x^3} \\
&= 3x + \frac{3}{x} \\
&= 3 \left( x + \frac{1}{x} \right).
\end{align*}By AM-GM, $x + \frac{1}{x} \ge 2,$ so
\[f(x) = 3 \left( x + \frac{1}{x} \right) \ge 6.\]Equality occurs at $x = 1,$ so the minimum value of $f(x)$ for $x > 0$ is $\boxed{6}.$