Putnam 1998.

Similar solution with @Hana Wehbi 's

$\begin{aligned} f(x) & = \frac {\left(x + \frac 1x\right)^6 - {\color{#3D99F6}\left(x^6 + \frac 1{x^6}\right)} - 2}{\left(x + \frac 1x\right)^3 + \left(x^3 + \frac 1{x^3}\right)} & \small \color{#3D99F6} \text{Note that }\left(x^3 + \frac 1{x^3}\right)^2 = x^6 + 2 + \frac 1{x^6} \\ & = \frac {\left(x + \frac 1x\right)^6 - {\color{#3D99F6}\left(x^3 + \frac 1{x^3}\right)^2+2} - 2}{\left(x + \frac 1x\right)^3 + \left(x^3 + \frac 1{x^3}\right)} \\ & = \frac {\left(x + \frac 1x\right)^6 - \left(x^3 + \frac 1{x^3}\right)^2}{\left(x + \frac 1x\right)^3 + \left(x^3 + \frac 1{x^3}\right)} \\ & = \frac {\left[\left(x + \frac 1x\right)^3 + \left(x^3 + \frac 1{x^3}\right)\right]\left[\left(x + \frac 1x\right)^3 - \left(x^3 + \frac 1{x^3}\right)\right]}{\left(x + \frac 1x\right)^3 + \left(x^3 + \frac 1{x^3}\right)} \\ & = \left(x + \frac 1x\right)^3 - \left(x^3 + \frac 1{x^3}\right) \\ & = 3 \left(\color{#3D99F6} x+\frac 1x\right) & \small \color{#3D99F6} \text{By AM-GM inequality: } x + \frac 1x \ge 2 \sqrt {x \times \frac 1x} = 2 \\ & \ge 3 \times {\color{#3D99F6} 2} = \boxed{6} & \small \color{#3D99F6} \text{Equality occurs when } x=1 \end{aligned}$

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Reference: AM-GM inequality

On expanding (x+1/x)^6 and (x+1/x)^3 and simplifying the given expression, we get the given expression equal to 3(x+1/x). Since the minimum of (x+1/x) for x>0 is 2, therefore the minimum of the given expression is (3)(2)=6

Let's start by expanding $f(x)$ :

$\large f(x)= \Large \frac{(x+\frac{1}{x})^6 -(x^6 +\frac{1}{x^6}) -2}{(x+\frac{1}{x})^3 +(x^3 +\frac{1}{x^3})}$ =

$\large f(x)= \Large \frac{(x+\frac{1}{x})^6 -(x^6 +\frac{1}{x^6} +2) }{(x+\frac{1}{x})^3 +(x^3 +\frac{1}{x^3})}$ =

$\large f(x)= \Large \frac{((x+\frac{1}{x})^3)^2 - (x^3 +\frac{1}{x^3})^2}{(x+\frac{1}{x})^3 +(x^3 +\frac{1}{x^3})}$ =

$\large f(x)= \Large \frac{((x+\frac{1}{x})^3 +(x^3 +\frac{1}{x^3})) \times ((x+\frac{1}{x})^3 - (x^3 +\frac{1}{x^3}) )}{(x+\frac{1}{x})^3 +(x^3 +\frac{1}{x^3})}$ =

$f(x)=(x+\frac{1}{x})^3 - (x^3 +\frac{1}{x^3})= 3x +\frac{3}{x}.$

$f'(x)= 3-\frac{3}{x^2} =0 \implies x=\pm 1.$

Since $x>0$ the critical point is at $x=1.$

$f"(x)=\frac{1}{x^3} \implies f"(1)=1 >0 \implies x=1$ gives a minimum at $f(1)= 3(1)+\frac{1}{1} =\boxed{6}$ since $f(x)=3x+\frac{1}{x}.$

$\frac{-\left(x^6+\frac{1}{x^6}\right)+\left(x+\frac{1}{x}\right)^6-2}{\left(x^3+\frac{1}{x^3}\right)+\left(x+\frac{1}{x}\right)^3} \Rightarrow 3-\frac{3}{x^2}$

$\frac{\partial \left(3 x+\frac{3}{x}\right)}{\partial x}=0 \Rightarrow x\to -1, x\to 1$

With the problem specified limitation on the value of x and using the simplified form, the answer is 6.