Can it be an integer?

Similar solution with @Chan Lye Lee

$\begin{aligned} (u+v)^3 & = u^3+v^3+3uv(u+v) \\ & = 4 + \frac {44}{5\sqrt 5} + 4 - \frac {44}{5\sqrt 5} + 3 \sqrt[3]{4 + \frac {44}{5\sqrt 5}} \sqrt[3]{4 - \frac {44}{5\sqrt 5}} (u+v) \\ & = 8 + 3 \sqrt[3]{\left(4 + \frac {44}{5\sqrt 5}\right) \left(4 - \frac {44}{5\sqrt 5}\right)} (u+v) \\ & = 8 + 3 \sqrt[3]{4^2 - \frac {44^2}{(5\sqrt 5)^2}} (u+v) \\ & = 8 + 3 \sqrt[3]{\frac {64}{125}} (u+v) \\ & = 8 + \frac {12}5 (u+v) & \small \color{#3D99F6} \text{Multiply both sides by 5} \\ 5(u+v)^3 & = 40 + 12(u+v) \end{aligned}$

$\implies 5(u+v)^3 - 12(u+v) = \boxed{40}$

We first note that $uv=\sqrt[3]{4+\frac{44}{5\sqrt{5}}} \times \sqrt[3]{4-\frac{44}{5\sqrt{5}}} =\sqrt[3]{4^2-\frac{44^2}{5^2 \times 5}}= \sqrt[3]{ \frac{4^2}{125} \left(125-11^2\right)} = \frac{4}{5}$ .

Next, $u^3=4+\frac{44}{5\sqrt{5}}$ and $v^3=4-\frac{44}{5\sqrt{5}}$ , which imply that $u^3+v^3=8$ .

Finally, $5\left(u+v\right)^3-12\left(u+v\right) = 5\left(u^3+3uv(u+v)+v^3\right)-12\left(u+v\right) = 5\left(8+3\times \frac{4}{5}(u+v)\right) -12\left(u+v\right)=40.$