<p>Find the value of <span>\((\sqrt{2} + 1)^6 - (\sqrt{2} - 1)^6\)</span>.</p>
Step-by-Step Solution
Key Concept: When subtracting two binomial expansions with opposite signs in one term, all even-powered terms cancel out, leaving only odd-powered terms with double coefficient. Use the binomial theorem systematically and exploit symmetry.
<p><strong>Step 1:</strong> Apply binomial theorem to both expressions:</p><p>$(\sqrt{2} + 1)^6 = \binom{6}{0}(\sqrt{2})^6 + \binom{6}{1}(\sqrt{2})^5 + \binom{6}{2}(\sqrt{2})^4 + \binom{6}{3}(\sqrt{2})^3 + \binom{6}{4}(\sqrt{2})^2 + \binom{6}{5}\sqrt{2} + \binom{6}{6}$</p><p>$(\sqrt{2} - 1)^6 = \binom{6}{0}(\sqrt{2})^6 - \binom{6}{1}(\sqrt{2})^5 + \binom{6}{2}(\sqrt{2})^4 - \binom{6}{3}(\sqrt{2})^3 + \binom{6}{4}(\sqrt{2})^2 - \binom{6}{5}\sqrt{2} + \binom{6}{6}$</p><p><strong>Step 2:</strong> Subtract the second from the first. Even-powered terms cancel:</p><p>$(\sqrt{2} + 1)^6 - (\sqrt{2} - 1)^6 = 2\left[\binom{6}{1}(\sqrt{2})^5 + \binom{6}{3}(\sqrt{2})^3 + \binom{6}{5}\sqrt{2}\right]$</p><p><strong>Step 3:</strong> Calculate each odd term:</p><p>$\binom{6}{1}(\sqrt{2})^5 = 6 \cdot 4\sqrt{2} = 24\sqrt{2}$</p><p>$\binom{6}{3}(\sqrt{2})^3 = 20 \cdot 2\sqrt{2} = 40\sqrt{2}$</p><p>$\binom{6}{5}\sqrt{2} = 6\sqrt{2}$</p><p><strong>Step 4:</strong> Sum and multiply by 2:</p><p>$2(24\sqrt{2} + 40\sqrt{2} + 6\sqrt{2}) = 2 \cdot 70\sqrt{2} = 140\sqrt{2}$</p><p>∴ Answer: $140\sqrt{2}$</p>
Correct Answer: 140√2