Đáp án:
`(x^3 - x^2)^2 - 4x^2 + 8x - 4 = 0`
`⇔ (x^3 - x^2)^2 - 4(x^2 - 2x + 1) = 0`
`⇔ (x^3 - x^2)^2 - 4(x - 1)^2 = 0`
`⇔ (x^3 - x^2)^2 - [2(x - 1)]^2 = 0`
`⇔ [(x^3 - x^2) - 2(x - 1)][(x^3 - x^2) + 2(x - 1)] = 0`
`⇔ [x^2 (x - 1) - 2(x - 1)][x^2 (x - 1) + 2(x - 1)] = 0`
`⇔ (x - 1)(x^2 - 2)(x - 1)(x^2 + 2) = 0`
`⇔ (x - 1)^2 (x^2 - 2)(x^2 + 2) = 0`
`⇒` $\left[\begin{matrix} (x - 1)^2 = 0\\x^2 - 2 = 0\\x^2 + 2 = 0\end{matrix}\right.$
`⇒` $\left[\begin{matrix} x - 1 = 0\\x^2 = 2\\x^2 = - 2 (loại vì x^2 ≥ 0)\end{matrix}\right.$
`⇒` $\left[\begin{matrix} x = 1\\x =\left[\begin{matrix} \sqrt{2}\\-\sqrt{2}\end{matrix}\right.\end{matrix}\right.$
Vậy `x ∈ {1,`$\sqrt{2}$ `; -`$\sqrt{2}$`}`
