Discussion
The coefficient of \(x^{8}\) in the expansion of \(\left(x^{2} + \sqrt{1 - x^{2}}\right)^{5}\) + \(\left(x^{2} - \sqrt{1 - x^{2}}\right)^{5}\) , is
Answer: A
The coefficient of \(x^{8}\) is -20.
With \(a = x^{2}\) and \(b = \sqrt{1 - x^{2}}\), the odd powers of \(b\) cancel in \((a + b)^{5} + (a - b)^{5}\):
\[ \begin{aligned} (a+b)^{5} + (a-b)^{5} &= 2(a^{5} + 10a^{3}b^{2} + 5ab^{4}) \\ &= 2[x^{10} + 10x^{6}(1 - x^{2}) \\ &\quad + 5x^{2}(1 - x^{2})^{2}] \end{aligned} \]
Only \(10x^{6}(1 - x^{2})\) produces \(x^{8}\), with coefficient \(-10\); the last term reaches only \(x^{6}\). Doubling gives \(-20\). Forgetting the factor 2 gives \(-10\).
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