jmc

Let $y=\sqrt[3]{2-x}.$ Then $y^3=2-x,$ so $x=2-y^3.$ Then $$\sqrt{x-1}=\sqrt{1-y^3},$$so we can write the given equation as $y+\sqrt{1-y^3}=1.$ Then $$\sqrt{1-y^3}=1-y.$$Squaring both sides, we get $1-y^3=1-2y+y^2,$ so $y^3+y^2-2y=0.$ This factors as $y(y-1)(y+2)=0,$ so $y$ can be 0, 1, or $-2.$ These lead to $x=\boxed{1,2,10}.$ We check that these solutions work.