Решите уравнение: корень 3 степени из (4x-4)=x-1.

\[\sqrt[3]{4x - 4} = x - 1\]

\[4x - 4 = (x - 1)^{3}\]

\[4x - 4 = x^{3} - 3x^{2} + 3x - 1\]

\[x^{3} - 3x^{2} + 3x - 1 - 4x + 4 = 0\]

\[x^{3} - 3x^{2} - x + 3 = 0\]

\[x^{2}(x - 3) - (x - 3) = 0\]

\[(x - 3)\left( x^{2} - 1 \right) = 0\]

\[(x - 3)(x - 1)(x + 1) = 0\]

\[x = 3;\ \ x = 1;\ \ x = - 1.\]

\[Ответ:x = - 1;\ \ x = 1;\ \ x = 3.\]