jmc

Let $$y=\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\cdots }}}.$$Then $$y^3=x\sqrt[3]{x\sqrt[3]{x\cdots }}=xy,$$so $y^2=x.$

Let $$z=\sqrt[3]{x+\sqrt[3]{x+\sqrt[3]{x+\cdots }}}.$$Then $$z^3=x+\sqrt[3]{x+\sqrt[3]{x+\cdots }}=x+z,$$so $z^3-z=x.$

Since $z=y,$ $y^3-y=x=y^2.$ Then $$y^3-y^2-y=0,$$which factors as $y(y^2-y-1)=0,$ so $y^2-y-1=0.$ By the quadratic formula, $$y=\frac{1\pm \sqrt{5}}{2}.$$Since $y$ is positive, $$y=\frac{1+\sqrt{5}}{2}.$$Then $$x=y^2=\boxed{\frac{3+\sqrt{5}}{2}}.$$