\sqrt[3]{2+\sqrt{5} }

=\sqrt[3]{\dfrac{8(2+\sqrt{5})}{8} }

Now Let the numerator part

Using the cube of binomial formula

(a+b)^3=a^3+3a^2b + 3ab^2 +b^3

(x+y\sqrt{5})^3 = x^3+3x^2y\sqrt{5}+3x(y\sqrt{5} )^2 +(y\sqrt{5} )^3

= x^3+3x^2y\sqrt{5}+15xy^2 +5y^2\sqrt{5}

=(x^3+15xy^2)+(3x^2y\sqrt{5}+5y^2\sqrt{5})

Comparing the equations (2) and (3) gives the following equation system

\begin{cases} x^3+15xy^2 &=16 \\ 3x^2y+5y^2 &=8 \end{cases}

Since x,y are integers, it's easy to solve the system

x=1, y=1

Substituting to (2), the expression (1) is transformed to

\sqrt[3]{\dfrac{16+8\sqrt{5} }{8}}

=\sqrt[3]{(\dfrac{1+\sqrt{5} }{2} )^3}

=\dfrac{1+\sqrt{5} }{2}