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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

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$$b=c+8$$

Substituting in the second equation we get

$$c(c+8)+a^2+16 = 0$$

$$c^2+8c+16+a^2 = 0$$

We can rewrite this as 

$$ (c+4)^2 +a^2 = 0$$

Sum of squares is 0 implies that both the squares must be 0

$$\implies a = 0, c = -4, b = 4$$

$$a^{2025} + b^{2025} + c^{2025} = 0^{2025} + (4)^{2025} + (-4)^{2025} = 0$$

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