JEE Math Practice Question
Let $$\alpha, \beta$$ be the roots of the quadratic equation $$x^{2}+\sqrt{6} x+3=0$$. Then $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$$ is equal to :
- A.72
- B.9
- C.729
- D.81
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