jmc

Let $y=\sqrt[3]{3x-2}.$ Then the given equation becomes $3y=x^3+2,$ so $$x^3=3y-2.$$Also, from the equation $y=\sqrt[3]{3x-2},$ $y^3=3x-2.$ Subtracting these equations, we get $$x^3-y^3=3y-3x.$$Then $x^3-y^3+3x-3y=0,$ which factors as $$(x-y)(x^2+xy+y^2+3)=0.$$Since $$x^2+xy+y^2+3={\left(x+\frac{y}{2}\right)}^2+\frac{3}{4}{y}^2+3>0,$$we must have $x=y.$ Then $x^3-3x+2=0,$ which factors as $(x-1{)}^2(x+2)=0.$ Hence, the solutions are $\boxed{1,-2}.$