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Quadratic Equations

VITEEE 2024

MCQ+4 / -12024

Given, $\frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}=k$, then $\frac{x^8+y^8}{x^8-y^8}$ is equal to

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