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VITEEE 2024
MCQ (Single Correct Answer)
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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
Questions Asked from Quadratic Equations (MCQ (Single Correct Answer))
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