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(A) Given the equation: $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$ Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 = 0$ Adding $2a^2b^2$ to both

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$45^\circ$ or $135^\circ$

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(A) Given the equation: $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$ Rearranging the terms: $a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 = 0$ Adding $2a^2b^2$ to both sides: $a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 + 2a^2b^2 = 2a^2b^2$ This simplifies to: $(a^2 + b^2 - c^2)^2 = 2a^2b^2$ Taking the square root: $a^2 + b^2 - c^2 = \pm \sqrt{2}ab$ Dividing by $2ab$: $\frac{a^2 + b^2 - c^2}{2ab} = \pm \frac{\sqrt{2}ab}{2ab} = \pm \frac{1}{\sqrt{2}}$ By the Law of Cosines,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$,so $\cos C = \pm \frac{1}{\sqrt{2}}$ Therefore,$C = 45^\circ$ or $C = 135^\circ$.

If $a, b$ and $c$ are the sides of a triangle such that $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$,then the angle opposite to the side $c$ is:

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