Solve it properly:

Dear Student,

Please find the below required solution.

$$\begin{gathered} Given that \\ x = \frac{1}{4-\sqrt{15}} ...\left(1\right) \\ And y = \frac{1}{4+\sqrt{15}} ...\left(2\right) \\ x=\frac{1}{4-\sqrt{15}}= \frac{1}{4-\sqrt{15}}\times \frac{4+\sqrt{15}}{4+\sqrt{15}} [ On rationali\sin g ] \\ x = \frac{4+\sqrt{15}}{16-15} [ \sin ce a^2-b^2 = (a+b\left)\right(a-b) ] \\ x = 4+\sqrt{15} ...\left(3\right) \\ And y=\frac{1}{4+\sqrt{15}}= \frac{1}{4+\sqrt{15}}\times \frac{4-\sqrt{15}}{4-\sqrt{15}} [ On rationali\sin g ] \\ y = \frac{4-\sqrt{15}}{16-15} [ \sin ce a^2-b^2 = (a+b\left)\right(a-b) ] \\ y = 4-\sqrt{15} ...\left(4\right) \\ So value of x^3+y^3 is \\ {x}^3+y^3 = (x+y{)}^3-3xy(x+y) \\ = {\left[4+\sqrt{15} +4-\sqrt{15} \right]}^3-3(4+\sqrt{15} \left)\right(4-\sqrt{15} )\left[4+\sqrt{15} +4-\sqrt{15} \right] [ u\sin g \left(3\right) and \left(4\right) ] \\ = (8{)}^3-3(16-15)\left(8\right) [ \sin ce a^2-b^2 = (a+b\left)\right(a-b) ] \\ = 512-24 \\ = 488 \end{gathered}$$

Hope this information will clear your doubts about given problem.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.   

Regards