If \(x = \frac{y}{2}\),evaluate\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\)

\(x = \frac{y}{2} \)

\(\left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right)\)

\(\frac{x^3}{y^3} + \frac{1}{2} = (\frac{y}{2})^{3} \div y^{3} + \frac{1}{2}\)

= \(\frac{y^{3}}{8} \times \frac{1}{y^3} + \frac{1}{2}\)

= \(\frac{1}{8} + \frac{1}{2}\)

= \(\frac{5}{8}\)

\(\frac{1}{2} - \frac{x^2}{y^2} = \frac{1}{2} - (\frac{y}{2})^{2} \div y^2)\)

= \(\frac{1}{2} - \frac{y^2}{4} \times \frac{1}{y^2}\)

= \(\frac{1}{2} - \frac{1}{4}\)

= \(\frac{1}{4}\)

\(\therefore \left(\frac{x^{3}}{y^{3}}+\frac{1}{2}\right) \div \left(\frac{1}{2} - \frac{x^{2}}{y^{2}}\right) = \frac{5}{8} \div \frac{1}{4}\)

= \(\frac{5}{2}\)

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