$$
\text { 1. Given } \sqrt[3]{a}+a=\sqrt{6}, b^{3}+b=\sqrt{6} \text {. Then } a+b
$$
Let $f(x)=x^{3}+x$.
Since $f^{\prime}(x)=3 x^{2}+1>0$, $f(x)$ is monotonically increasing on $(-\infty,+\infty)$.
Given that $f(\sqrt[3]{a})=f(b)$.
Therefore, $\sqrt[3]{a}=b \Rightarrow a=b^{3}$.
Thus, $a+b=b^{3}+b=\sqrt{6}$.