If$$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\ $$for x > 0, then what is the value of $$\ x^{3}+\frac{1}{x^{3}}\ $$

Given : $$\ x^{2}+\frac{1}{x^{2}}=\frac{7}{4}\ $$

=> $$(x+\frac{1}{x})^2-2(x)(\frac{1}{x})=\frac{7}{4}$$

=> $$(x+\frac{1}{x})^2=\frac{7}{4}+2=\frac{15}{4}$$

=> $$x+\frac{1}{x}=\frac{\sqrt{15}}{2}$$

Cubing both sides, we get :

=> $$x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=(\frac{\sqrt{15}}{2})^3$$

=> $$x^3+\frac{1}{x^3}+3(\frac{\sqrt{15}}{2})=\frac{15\sqrt{15}}{8}$$

=> $$x^3+\frac{1}{x^3}=\frac{15\sqrt{15}}{8}-\frac{3\sqrt{15}}{2}$$

=> $$x^3+\frac{1}{x^3}=\frac{3\sqrt{15}}{8}$$

=> Ans - (C)