System of equations

Let $\frac{1}{\sqrt[3]{x}}=a$ and $\frac{1}{\sqrt[3]{y}}=b$ , then the system of equations become:

$a^3+b^3=9 \implies \left(a+b\right)^3-3ab\left(a+b\right)=9$

$\left(a+b\right)\left(1+a\right)\left(1+b\right)=18 \implies \left(a+b\right)\left(1+\left(a+b\right)+ab\right)=18$

Let $a+b=c$ and $ab=d$ , then the system of equations reduce to:

$c^3-3cd=9$ (I) and $c\left(1+c+d\right)=18 \implies c+c^2+cd=18 \implies 3c+3c^2+3cd=54$ (II)

Adding (I) and (II) :

$c^3+3c^2+3c=63 \implies c^3+3c^2+3c+1=64 \implies \left(c+1\right)^3=64 \implies c+1 = 4 \implies \boxed{c=3} \implies \boxed{a+b=3}$

Substituting $c = 3$ in (I) , we have $\boxed{d=2} \implies \boxed{ab=2}$ .

Solving $a+b=3$ and $ab=2$ , we have $a=1$ , $b=2$ or $b=1$ , $a=2$ .

Which means that $\frac{1}{\sqrt[3]{x}}=1$ , $\frac{1}{\sqrt[3]{y}}=2$ or $\frac{1}{\sqrt[3]{y}}=1$ , $\frac{1}{\sqrt[3]{x}}=2$ .

Subsequently, $x = 1$ , $y=\frac{1}{8}$ or $y=1$ , $x=\frac{1}{8}$ .

Therefore, $x+y = \frac{9}{8} = \boxed{1.125}$ in either case.

Using the identity

$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$

$\left(\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}\qquad}+\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}}+1\right)^3=\dfrac{1}{x}+\dfrac{1}{y}+1+3\left(\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}}+\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}}\right)\left(\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}}+1\right)\left(1+\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}}\right)=9+1+3(18)=10+54=64$

Hence,

$\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}} + \dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}} + 1=4$

It follows that

$\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}} + \dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}} =3$

The system is now reduced to

$\dfrac{1}{x}+\dfrac{1}{y}=9$

$\dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x}} + \dfrac{1}{\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{y}} =3$

Which is a symmetric system, having the solution

$x=\dfrac{1}{8},y=1$ and $x=1,y=\dfrac{1}{8}$

Finally,

$x+y=\dfrac{1}{8}+1=\dfrac{9}{8}=$ $\color{#D61F06}\large \boxed{1.125}$