An algebra problem by Lucas Tan

$\begin{aligned} \sqrt[3]{\sqrt{a^3+b^2}+b} \times\sqrt[3]{\sqrt{a^3+b^2}-b} & = 1 - 2a \\ \sqrt[3]{\color{#3D99F6}\left(\sqrt{a^3+b^2}+b \right) \left(\sqrt{a^3+b^2}-b\right)} & = 1 - 2a & \small \color{#3D99F6} \text{Note that }(x+y)(x-y) = x^2 - y^2 \\ \sqrt[3]{a^3+b^2 - b^2} & = 1 - 2a \\ \sqrt[3]{a^3} & = 1 - 2a \\ a & = 1 - 2a \\ 3a & = 1 \\ \implies a & = \frac 13 \approx \boxed{0.333} \end{aligned}$

I first cubed both sides and eliminated b from the equation as @Chew-Seong Cheong did to obtain the following

$a^3 = ( 1 -2a)^3$

Using the expansion of $(x + y)^3$ , I solved it further to get

$(3a - 1)(3a^2 - 3a + 1) = 0$ Solving this gives $a = 1/3$ and $a = 1/2 ± i√3/6$ .

Hence there are 3 solutions, one real and two unreal.