\sqrt[4]{x + 1} = 0 \\= > x_1= - 1 \\ \\ 2 {y}^{3} - 1 = 0 \\ {y}^{3} = \frac{1}{2} \\ y_2= \sqrt[3]{ \frac{1}{2} } \\ \sqrt[4]{x + 1} = \sqrt[3]{ \frac{1}{2} } \\ x + 1 = \frac{1}{2} \sqrt[3]{ \frac{1}{2} } \\ x_2= \frac{1}{2} \sqrt[3]{ \frac{1}{2} } - 1 \\ \\ \\ x_1= - 1\\ x_2= \frac{1}{2} \sqrt[3]{ \frac{1}{2} } - 1 \\ " alt=" \sqrt[4]{ x + 1} = 2x + 2\\ \sqrt[4]{ x + 1} = 2(x + 1) \\ y = \sqrt[4]{x + 1} \geqslant 0 \\ y = 2 {y}^{4} \\ 2 {y}^{4} - y = 0 \\ y(2 {y}^{3} - 1) = 0 \\ y_1= 0 = > \sqrt[4]{x + 1} = 0 \\= > x_1= - 1 \\ \\ 2 {y}^{3} - 1 = 0 \\ {y}^{3} = \frac{1}{2} \\ y_2= \sqrt[3]{ \frac{1}{2} } \\ \sqrt[4]{x + 1} = \sqrt[3]{ \frac{1}{2} } \\ x + 1 = \frac{1}{2} \sqrt[3]{ \frac{1}{2} } \\ x_2= \frac{1}{2} \sqrt[3]{ \frac{1}{2} } - 1 \\ \\ \\ x_1= - 1\\ x_2= \frac{1}{2} \sqrt[3]{ \frac{1}{2} } - 1 \\ " align="absmiddle" class="latex-formula">