If x^2 + 1x^2 = 7/4 for x > 0 then what is the value of x^3 + 1x^3 = ?

Answer: Option C

$$\begin{array}{rl}& \text{Given, }{x}^2+\frac{1}{x^2}=\frac{7}{4}\\ & \text{Adding 2 to both sides}\\ & x^2+\frac{1}{x^2}+2=\frac{7}{4}+2\\ & {\left(x+\frac{1}{x}\right)}^2=\frac{15}{4}\\ & x+\frac{1}{x}=\frac{\sqrt{15}}{2}\\ & \because x^3+\frac{1}{x^3}={\left(x+\frac{1}{x}\right)}^3-3\left(x+\frac{1}{x}\right)\\ & \therefore x^3+\frac{1}{x^3}=\frac{15\times \sqrt{15}}{8}-3\times \frac{\sqrt{15}}{2}\\ & =\frac{15\sqrt{15}-12\sqrt{15}}{8}\\ & =\frac{3\sqrt{15}}{8}\end{array}$$