help

Let  $a^2=\frac{16}{44}$  and  $b^2=\frac{(2+\sqrt{5})^2}{11}$ , where  $a$  is a negative real number and  $b$  is a positive real number.

If  $(a+b)^3$  can be expressed in the simplified form  $\frac{x\sqrt{y}}{z}$  where  $x$,  $y$,  and  $z$  are positive integers,

what is the value of the sum  $\vphantom{\frac{x\sqrt{y}}{z}}x+y+z$ ?

______________________________________

$a\quad=\quad-\sqrt{\frac{16}{44}}\quad=\quad -\frac{4}{2\sqrt{11}}\quad=\quad\frac{-2}{\sqrt{11}}\\~\\ b\quad=\quad\sqrt{\frac{(2+\sqrt{5})^2}{11}}\quad=\quad\frac{2+\sqrt{5}}{\sqrt{11}}\\~\\~\\ (a+b)^3\,=\,\Big(\frac{-2}{\sqrt{11}}+\frac{2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{-2+2+\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\,\Big(\frac{\sqrt{5}}{\sqrt{11}}\Big)^3\\~\\ (a+b)^3\,=\, \frac{\sqrt{5}}{\sqrt{11}} \cdot\frac{\sqrt{5}}{\sqrt{11}}\cdot\frac{\sqrt{5}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{5}}{11\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}} \\~\\ (a+b)^3\,=\, \frac{5\sqrt{55}}{121}$

Now it is in the form  $\frac{x\sqrt{y}}{z}$  where  x,  y,  and  z  are positive integers.

x + y + z  =  5 + 55 + 121  =  181