Question

$\sqrt[3]{x + 1} = 1 + 2x$
x=?​

Answer 1

$\sqrt[3]{x + 1} = 1 + 2x \: | {( \: )}^{3}$

$x + 1 = {1}^{3} + 3 \times {1}^{2} \times 2x + 3 \times 1 \times {(2x)}^{2} + {(2x)}^{3}$

$x + 1 = 1 + 6x + 3 \times 4 {x}^{2} + 8 {x}^{3}$

$x + 1 = 1 + 6x + 12 {x}^{2} + 8 {x}^{3}$

$8 {x}^{3} + 12 {x}^{2} + 6x + 1 - x - 1 = 0$

$8 {x}^{3} + 12 {x}^{2} + 5x = 0$

$x(8 {x}^{2} + 12x + 5) = 0$

$x_{1} = 0$

$8 {x}^{2} + 12x + 5 = 0$

$a = 8$

$b = 12$

$c = 5$

$\Delta = {b}^{2} - 4ac$

$\Delta = {12}^{2} - 4 \times 8 \times 5$

$\Delta = 144 - 160$

$\Delta = - 16$

$x_{2,3}=\frac{-b \pm i\sqrt{-\Delta}}{2a} = \frac{ - 12 \pm i\sqrt{ - ( - 16)} }{2 \times 8} = \frac{ - 12 \pm i \sqrt{16} }{16} = \frac{ - 12 \pm4i}{16}$

$x_{2} = \frac{ - 12 + 4i}{16} = \frac{4( - 3 + i)}{16} = \frac{ - 3 + i}{4}$

$x_{3} = \frac{ - 12 - 4i}{16} = \frac{4( - 3 - i)}{16} = \frac{ - 3 - i}{4}$