Rational Function Minimization 2
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 $.
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
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- $\sqrt{}$
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Solution
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} $.
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