$\sqrt[3]{x+1} + \sqrt[3]{x-1} = x \sqrt[3]{2}$
Đặt $\sqrt[3]{x+1} = a$ ; $\sqrt[3]{x-1}=b$
\Rightarrow $a^3-b^3 = 2$ ; $a^3+b^3 = 2x$
PT \Leftrightarrow $a+b=\dfrac{a^3+b^3}{2}.\sqrt[3]{2}$
\Leftrightarrow $a+b=\dfrac{(a+b)(a^2-ab+b^2)}{2}.\sqrt[3]{2}$
\Leftrightarrow a+b = 0 or $1=\dfrac{a^2-ab+b^2}{2}.\sqrt[3]{2}$
*TH1: a+b = 0 \Leftrightarrow a = -b \Leftrightarrow .....
*TH2: $1=\dfrac{a^2-ab+b^2}{2}.\sqrt[3]{2}$