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

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

\[4x + 4 = (x + 1)³\]

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

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

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

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

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

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

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

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