Question

If \(x>0\), then the minimum value of \[ \dfrac{\left(x+\dfrac{1}{x}\right)^6 - \left(x^6+\dfrac{1}{x^6}\right) - 2}{\left(x+\dfrac{1}{x}\right)^3 + \left(x^3+\dfrac{1}{x^3}\right)} \] is:

• A. 6
• B. 3
• C. 2
• D. 1

Hint:

Try writing \(x^6+\dfrac{1}{x^6}\) in terms of \(x^3+\dfrac{1}{x^3}\), the same way \(x^2+\dfrac{1}{x^2}\) can be written in terms of \(x+\dfrac{1}{x}\).

Answer 1

The Correct Option is A

Step 1: Spot a difference-of-squares pattern instead of a two-stage substitution.
Let $u = x+\frac{1}{x}$ and $v = x^3+\frac{1}{x^3}$. Squaring $v$: $v^2 = x^6+\frac{1}{x^6}+2$, so $x^6+\frac{1}{x^6} = v^2-2$.

Step 2: Rewrite the numerator as $u^6-v^2$.
The numerator is $u^6-\left(x^6+\frac{1}{x^6}\right)-2 = u^6-(v^2-2)-2 = u^6-v^2$.

Step 3: Factor using $A^2-B^2=(A-B)(A+B)$.
Treat $u^6$ as $(u^3)^2$, so $u^6-v^2 = (u^3-v)(u^3+v)$.

Step 4: Match the denominator to one of the factors.
The given denominator is $u^3+v$, exactly the second factor above. So the whole fraction collapses to $$\dfrac{(u^3-v)(u^3+v)}{u^3+v} = u^3-v$$ as long as $u^3+v\neq0$, which holds here since both $u$ and $v$ are positive for $x>0$.

Step 5: Simplify $u^3-v$.
Since $u^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right) = v+3u$, we get $u^3-v = 3u = 3\left(x+\frac{1}{x}\right)$, the same simplified form found the other way, but reached here purely by factoring instead of a two-step substitution chain.

Step 6: Apply AM-GM.
For $x>0$, AM-GM gives $x+\frac{1}{x}\geq2$, with equality at $x=1$. So $3u\geq6$, and the minimum value $6$ is reached at $x=1$.

Final Answer:
The minimum value of the expression is $6$. $$\boxed{6}$$