ĐKXĐ : `x≥1`

$\sqrt[3]{x+6}+\sqrt{x-1}=x^2 -1$

$⇔\sqrt[3]{x+6}-2+\sqrt{x-1}-1=x^2-4$

$⇔\dfrac{x+6-8}{(\sqrt[3]{x+6})^2+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt{x-1}+1}=x^2-4$

$⇔\dfrac{x-2}{(\sqrt[3]{x+6})^2+2\sqrt[3]{x+6}+4}+\dfrac{x-2}{\sqrt{x-1}+1}-(x-2)(x+2)=0$

$⇔(x-2)(\dfrac{1}{\sqrt[3]{(x+6)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt{x-1}+1}-x-2)=0$  `(1)`

`∀x≥1` ta có : $\dfrac{1}{\sqrt[3]{(x+6)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt{x-1}+1}>0$

`-(x+2)≤-3`

$⇒\dfrac{1}{\sqrt[3]{(x+6)^2}+2\sqrt[3]{x+6}+4}+\dfrac{1}{\sqrt{x-1}+1}-x-2\ne0$ `(2)`

Từ `(1)` và `(2)⇒x-2=0`

`⇔x=2`  `(` TMĐK `)`

Vậy `S={2}`