Question 16
$$a, b, c$$ are real numbers such that $$b - c = 8$$ and $$bc + a^2 + 16 = 0$$.
The numerical value of $$a^{2025} + b^{2025} + c^{2025}$$ is ------.
Correct Answer: 0
Solution
Since $$(b+c)^2 = (b-c)^2+4bc = 64+4(-a^2-16) = -4a^2$$, and a square cannot be negative, we need $$a=0$$, forcing $$b+c=0$$ as well. Combined with $$b-c=8$$, this gives $$b=4, c=-4$$. Then $$a^{2025}+b^{2025}+c^{2025} = 0+4^{2025}+(-4)^{2025} = 0$$, since $$2025$$ is odd.
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