For each positive number $x$, let 
$f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$. 
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$
Let $a = \left( x + \frac{1}{x} \right)^3$ and $b = x^3 + \frac{1}{x^3}$. Then
\begin{align*} f(x) &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + 2 + \frac{1}{x^6})}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^3 + \frac{1}{x^3})^2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{a^2 - b^2}{a + b}. \end{align*}
By difference of squares,
\begin{align*} f(x) &= \frac{(a - b)(a + b)}{a + b} \\ &= a - b \\ &= \left( x + \frac{1}{x} \right)^3 - \left( x^3 + \frac{1}{x^3} \right) \\ &= \left( x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \right) - \left( x^3 + \frac{1}{x^3} \right) \\ &= 3x + \frac{3}{x} \\ &= 3 \left( x + \frac{1}{x} \right). \end{align*}
By the AM-GM inequality,
\[x + \frac{1}{x} \ge 2,\]
so $f(x) \ge 6$. Furthermore, when $x = 1$, $f(1) = 6$, so the minimum value of $f(x)$ is $\boxed{6}$.