Cách khác nhanh hơn:
ĐK: $x \ge \sqrt[3]{2}$
$PT \leftrightarrow (\sqrt[3]{x^2 - 1} - 2) + (x - 3) = \sqrt{x^3 - 2} - 5$
$\leftrightarrow \dfrac{x^2 - 9}{\sqrt[3]{(x^2 - 1)^2 } + 4 + 2\sqrt[3]{x^2 -1}} + (x-3) = \dfrac{x^3 - 27}{\sqrt{x^3 - 2} + 5}$
$\leftrightarrow (x-3)\underbrace{[\dfrac{x+3}{\sqrt[3]{(x^2 - 1)^2} + 4 + 2\sqrt[3]{x^2-1}} + 1 - \dfrac{x^2 + 3x + 9}{\sqrt{x^3 - 2} + 5}]} = 0$
...............................................<0,\forall $x \ge \sqrt[3]{2}$
$\leftrightarrow x - 3 = 0 \leftrightarrow x = 3 (thỏa)$