<p>319. If \(x > 0\) then the minimum value of \(x^3 + \frac{1}{x^3}\) is</p>
Step-by-Step Solution
Key Concept: Use AM-GM inequality on the expression by recognizing that x³ + 1/x³ can be bounded below using the identity (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x), combined with the minimum of (x + 1/x) which is 2 when x > 0.
<p><strong>Step 1:</strong> Let t = x + 1/x. Since x > 0, by AM-GM: t = x + 1/x ≥ 2√(x · 1/x) = 2, with equality when x = 1.</p><p><strong>Step 2:</strong> Cube the relation: t³ = (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x) = x³ + 1/x³ + 3t</p><p><strong>Step 3:</strong> Therefore: x³ + 1/x³ = t³ - 3t</p><p><strong>Step 4:</strong> Since t ≥ 2, let f(t) = t³ - 3t for t ≥ 2. Taking derivative: f'(t) = 3t² - 3 = 3(t² - 1) > 0 for t > 1. So f(t) is increasing on [2, ∞).</p><p><strong>Step 5:</strong> Minimum occurs at t = 2: f(2) = 8 - 6 = 2</p><p>∴ Minimum value of x³ + 1/x³ is <strong>2</strong> (achieved when x = 1)</p>
Correct Answer: C