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: Name a variable to simplify the expression.
Let \(a = x+\dfrac{1}{x}\). We will rewrite both the numerator and the denominator in terms of \(a\) alone, since \(x\) only appears through this combination.

Step 2: Express \(x^3+\dfrac{1}{x^3}\) using \(a\).
Cubing \(a\): \[ a^3 = \left(x+\dfrac{1}{x}\right)^3 = x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right) = \left(x^3+\dfrac{1}{x^3}\right)+3a \] So \(x^3+\dfrac{1}{x^3} = a^3-3a\). This means the denominator becomes \[ a^3 + (a^3-3a) = 2a^3-3a \]

Step 3: Express \(x^6+\dfrac{1}{x^6}\) using \(b=x^2+\dfrac{1}{x^2}\).
First, \(a^2 = x^2+\dfrac{1}{x^2}+2\), so \(b=a^2-2\). The same cubing trick used in Step 2, applied to \(b\) instead of \(a\), gives \(x^6+\dfrac{1}{x^6} = b^3-3b\) (since \(x^6=(x^2)^3\)). Substituting \(b=a^2-2\): \[ b^3-3b = (a^2-2)^3 - 3(a^2-2) = a^6-6a^4+12a^2-8-3a^2+6 = a^6-6a^4+9a^2-2 \]

Step 4: Build the numerator.
The numerator is \(a^6 - (x^6+\frac{1}{x^6}) - 2\), so \[ a^6 - (a^6-6a^4+9a^2-2) - 2 = 6a^4-9a^2 = 3a^2(2a^2-3) \]

Step 5: Divide numerator by denominator.
From Step 2, the denominator is \(2a^3-3a=a(2a^2-3)\). So \[ \dfrac{3a^2(2a^2-3)}{a(2a^2-3)} = 3a \] since \(a\neq0\) (as \(a=x+\frac{1}{x}\geq2>0\) for \(x>0\)) and \(2a^2-3\) cancels. So the whole expression is simply \(3\left(x+\dfrac{1}{x}\right)\).

Step 6: Minimize \(3\left(x+\dfrac{1}{x}\right)\) for \(x>0\).
By AM-GM, for \(x>0\), \(x+\dfrac{1}{x}\geq2\sqrt{x\cdot\dfrac{1}{x}}=2\), with equality exactly when \(x=\dfrac{1}{x}\), i.e. \(x=1\). So the minimum of \(3\left(x+\dfrac{1}{x}\right)\) is \(3\times2=6\), attained at \(x=1\).

Final Answer:
The minimum value of the given expression is \(6\). \[ \boxed{6} \]