If x^3+\frac{1}{x^3}=\frac{65}{8} and y^3+\frac{1}{y^3}=\frac{730}{27}, then which one of the following is a value of xy?

Explanation

We can rewrite the equations as x^3 + \frac{1}{x^3} = 8 + \frac{1}{8} and y^3 + \frac{1}{y^3} = 27 + \frac{1}{27}. From this, we deduce that x^3 = 8 \implies x = 2 and y^3 = 27 \implies y = 3. Therefore, a valid value for xy is 2 \times 3 = 6.

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