<p>The smallest integer larger than \((\sqrt{3} + \sqrt{2})^6\) is</p>
Step-by-Step Solution
Key Concept: Use binomial expansion to find that (√3 + √2)⁶ is slightly less than an integer by comparing it with (√3 - √2)⁶, which is a small positive number less than 1.
<p><strong>Step 1:</strong> Let α = (√3 + √2)⁶ and β = (√3 - √2)⁶</p><p><strong>Step 2:</strong> By binomial theorem, α + β contains only even-powered terms (all irrational parts cancel). Expanding: α + β = 2[C(6,0)·(√3)⁶ + C(6,2)·(√3)⁴·(√2)² + C(6,4)·(√3)²·(√2)⁴ + C(6,6)·(√2)⁶]</p><p><strong>Step 3:</strong> Calculate: α + β = 2[729 + 15·81·2 + 15·9·4 + 8] = 2[729 + 2430 + 540 + 8] = 2(3707) = 7414</p><p><strong>Step 4:</strong> Since 0 < √3 - √2 < 1 (≈ 0.318), we have 0 < β = (√3 - √2)⁶ < 1</p><p><strong>Step 5:</strong> Therefore α = 7414 - β where 0 < β < 1, so 7413 < α < 7414</p><p><strong>Step 6:</strong> The smallest integer larger than α is ⌈α⌉ = 7414</p><p>∴ Answer: 7414</p>
Correct Answer: C