\[\sqrt[6]{x^{2} + 4x - 1} = \sqrt[6]{x + 3}\]
\[x^{2} + 4x - 1 = x + 3\]
\[x^{2} + 3x - 4 = 0\]
\[x_{1} + x_{2} = - 3;\ \ \ x_{1} \cdot x_{2} = - 4\]
\[x_{1} = - 4;\ \ x_{2} = 1.\]
\[Проверка.\]
\[x = - 4:\]
\[\sqrt[6]{16 - 16 - 1} = \sqrt[6]{- 4 + 3}\]
\[\sqrt[6]{- 1} = \sqrt[6]{- 1}\]
\[не\ существует;\]
\[x = - 4\ не\ является\ корнем\ \]
\[уравнения.\]
\[x = 1:\]
\[\sqrt[6]{1 + 4 - 1} = \sqrt[6]{1 + 3}\]
\[\sqrt[6]{4} = \sqrt[6]{4}\]
\[x = 1 - корень\ уравнения.\]
\[Ответ:x = 1.\]