Решите уравнение: корень 3 степени из (13+x)=x+1.

\[\sqrt[3]{13 + x} = x + 1\]

\[13x + 1 = (x + 1)^{3}\]

\[13x + 1 = x^{3} + 3x^{2} + 3x + 1\]

\[- x^{3} - 3x^{2} + 10x = 0\]

\[x^{3} + 3x^{2} - 10x = 0\]

\[x\left( x^{2} + 3x - 10 \right) = 0\]

\[x(x + 5)(x - 2) = 0\]

\[x = 0;\ \ x = - 5;\ \ x = 2\]

\[Ответ:x = 0;\ \ x = - 5;\ \ x = 2.\]