let $\sqrt[3]{x} = t > 0$,then $\sqrt{1+\sqrt {1+t^3}}=t \rightarrow t^2-1 = \sqrt{1+t^3}$ $(t-1)^2(t+1)^2 = t^3 +1 = (t+1)(t^2-t+1)$ because $t+1 > 1$, we can cancel the terms on the two sides. $(t-1)^2(t+1) = (t^2-t+1)= (t-1)^2+t \rightarrow (t-1)^2(t) = t \rightarrow (t-1)^2 = 1 \rightarrow t = 2$, given $t >0$