Algebra - cube roots and square roots

$\begin{aligned} y & = \sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}} & \small \color{#3D99F6} \text{Cubing both sides.} \\ y^3 & = x+\sqrt{x^2+1} + 3\left(\sqrt[3]{x+\sqrt{x^2+1}}\right)^2 \sqrt[3]{x-\sqrt{x^2+1}} + 3\sqrt[3]{x+\sqrt{x^2+1}}\left(\sqrt[3]{x-\sqrt{x^2+1}}\right)^2 + x-\sqrt{x^2+1} \\ & = 2x + 3\sqrt[3]{(x+\sqrt{x^2+1})(x-\sqrt{x^2+1})} \left(\color{#3D99F6}\sqrt[3]{x+\sqrt{x^2+1}} + \sqrt[3]{x-\sqrt{x^2+1}}\right) \\ & = 2x + 3\sqrt[3]{-1}\color{#3D99F6} y \\ & = 2x - 3y \end{aligned}$

$\implies x = \dfrac {y^3+3y}2 = \boxed{\dfrac {y(y^2+3)}2}$

Let $A=\sqrt {x^2+1}\Rightarrow A ^2=x^2+ 1$ ,also we have that $(a+b)^3 = a^3+b^3+3ab(a+b)$ then our expression becomes $\begin{aligned} y & ={\color{#20A900} \sqrt[3]{x+A } + \sqrt[3]{x-A}}\\ y^3 & =2x +3\sqrt[3]{(x+A)(x-A)}\left({\color{#20A900}\sqrt [3]{x+A} + \sqrt[3]{x-A}}\right) \\ y^3& = 2x +3y \sqrt[3]{x^2-A^2}\end{aligned}$ Undo the substitution of $A^2$ and simplying we have $y^3 = 2x- 3y \implies x=\dfrac{y(y^2+3)}{2}$