Let's find the square root of 9+4√5. Let's suppose it has the form a+b√5, then (a+b√5)^2=a^2+5b^2+2ab√5=9+4√5. So a^2+5b^2=9 and ab=2, b=2/a.
Therefore a^2+5*4/a^2=9; multiply through by a^2: a^4+20-9a^2=0=(a^2-4)(a^2-5)=(a-2)(a+2)(a-√5)(a+√5)
We need a and b to be rational, so a=2 and b=1 or a=-2 and b=-1. The square root of 9+4√5 is therefore ±(2+√5).
We now know √y=2+√5 (assuming the positive square root). Let x=√y then x^3+1/x^3 is the original expression.
x^3=(2+√5)^3=8+3*4√5+3*2*5+5√5=38+17√5;
1/x^3=1/(38+17√5)=(38-17√5)/(38^2-5*17^2)=
(38-17√5)/(-1)=17√5-38=0.0131556 approx.
[38^2=1444, (17√5)^2=289*5=1445.]