Решите уравнение: корень 3 степени из (7x+1)=x+1.

\[\sqrt[3]{7x + 1} = x + 1\]

\[7x + 1 = (x + 1)^{3}\]

\[x^{3} + 3x^{2} + 3x - 7x + 1 - 1 = 0\]

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

\[x\left( x^{2} + 3x - 4 \right) = 0\]

\[x(x + 4)(x - 1) = 0\]

\[x = 0;\ \ x = - 4;\ \ x = 1.\]

\[Ответ:x = 0;\ \ x = - 4;\ \ x = 1.\]