$\sqrt[3]{x+1} + \sqrt[3]{x-1} = \sqrt[3]{5x}$ (ĐK: $x \in \mathbb{R}$)

$\iff (\sqrt[3]{x+1} + \sqrt[3]{x-1})^3 = (\sqrt[3]{5x})^3$

$\iff (x+1) + (x-1) + 3\sqrt[3]{x+1} \cdot \sqrt[3]{x-1} \cdot (\sqrt[3]{x+1} + \sqrt[3]{x-1}) = 5x$

$\iff 2x + 3\sqrt[3]{(x+1)(x-1)} \cdot \sqrt[3]{5x} = 5x$

$\iff 3\sqrt[3]{5x(x^2 - 1)} = 3x$

$\iff \sqrt[3]{5x(x^2 - 1)} = x$

$\iff 5x(x^2 - 1) = x^3$

$\iff 5x^3 - 5x = x^3$

$\iff 4x^3 - 5x = 0$

$\iff x(4x^2 - 5) = 0$

$\text{TH1:}$ $x = 0$

$\Rightarrow \begin{cases} \text{VT} = \sqrt[3]{0+1} + \sqrt[3]{0-1} = 1 + (-1) = 0\\\text{VP} = \sqrt[3]{5 \cdot 0} = 0 \end{cases}$

`=> \text{VT} = \text{VP}`

`=> x = 0` là nghiệm thoả mãn

$\text{TH2:}$ $4x^2 - 5 = 0$

$4x^2 = 5$

$x^2 = \dfrac{5}{4}$

$x = \pm \dfrac{\sqrt{5}}{2}$

`@` Với $x = \dfrac{\sqrt{5}}{2}$

$\Rightarrow \begin{cases} \text{VT} = \sqrt[3]{\dfrac{\sqrt{5}}{2}+1} + \sqrt[3]{\dfrac{\sqrt{5}}{2}-1} = \sqrt[3]{\dfrac{5\sqrt{5}}{2}}\\\text{VP} = \sqrt[3]{5 \cdot \dfrac{\sqrt{5}}{2}} = \sqrt[3]{\dfrac{5\sqrt{5}}{2}} \end{cases}$

$\Rightarrow \text{VT} = \text{VP}$

$\Rightarrow x = \dfrac{\sqrt{5}}{2}$ là nghiệm thỏa mãn

`@` Với $x = -\dfrac{\sqrt{5}}{2}$

$\Rightarrow \text{VT} = \text{VP} = -\sqrt[3]{\dfrac{5\sqrt{5}}{2}} \text{(TM)}$ 

Vậy $S = \left\{0; \, -\dfrac{\sqrt{5}}{2}; \, \dfrac{\sqrt{5}}{2}\right\}$

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