<p>The sum of the coefficients of all odd degree terms in the expansion of \(\left(x + \sqrt{x^3 - 1}\right)^5 + \left(x - \sqrt{x^3 - 1}\right)^5\), \((x > 1)\) is</p>
Step-by-Step Solution
Key Concept: When you add two binomial expansions with opposite middle terms, all odd-degree terms cancel out due to symmetry. Substituting x=1 directly into the simplified even-degree expression gives the sum of coefficients of odd-degree terms in the original expansion.
<p><strong>Step 1:</strong> Let $f(x) = (x + \sqrt{x^3-1})^5 + (x - \sqrt{x^3-1})^5$</p><p><strong>Step 2:</strong> Using binomial theorem, when we expand both terms and add them, all terms with odd powers of $\sqrt{x^3-1}$ cancel (since they have opposite signs). Only even-powered terms survive.</p><p><strong>Step 3:</strong> The result contains only even powers of $\sqrt{x^3-1}$, which when expanded gives only even-degree terms in $x$. The odd-degree terms in the original expression have coefficients that sum to zero.</p><p><strong>Step 4:</strong> To find the sum of coefficients of odd-degree terms, substitute $x=1$:<br>$f(1) = (1 + \sqrt{1-1})^5 + (1 - \sqrt{1-1})^5 = 1^5 + 1^5 = 2$</p><p><strong>Step 5:</strong> Since $f(1) = 2$ represents only the even-degree terms (odd-degree terms are already zero), and using the property that $f(1) + f(-1) = 2\times$(sum of even coefficients), we find the sum of coefficients of odd-degree terms = $\boxed{0}$</p><p>∴ Answer: A</p>
Correct Answer: A