Find Other $\scbrt{a}$ from Known $\scbrt{a}$

Pub: Sep. 14, 2014 | Wri: a few years ago

\[ \text{Let } x = \scbrt{a} \\ x^3 = a \\ x^3 - a = 0 \\ x^3 - \b{\scbrt{a}}^3 = 0 \\ \b{x - \scbrt{a}}\!\!\b{x^2 + x\scbrt{a} + \scbrt{a^2}} = 0 \\ x = \scbrt{a} \qquad x^2 + x\scbrt{a} + \scbrt{a^2} = 0 \\ % use \phantom to make x = \scbrt{a} top aligned \p{x = \scbrt{a}} \qquad \begin{split} \! x &= \f{-\scbrt{a} \pm \sqrt{\scbrt{a^2} - 4\scbrt{a^2}}}{2} \\[3pt] &= \f{-\scbrt{a} \pm \sqrt{-3\scbrt{a^2}}}{2} \\[3pt] &= \f{-\scbrt{a} \pm \scbrt{a}\sqrt{-3}}{2} \\[3pt] &= \scbrt{a}\b{-\f{1}{2} \pm \f{\sqrt{3}}{2}i} \end{split} \\[63pt] x_1 = \scbrt{a} \qquad x_2 = \scbrt{a}\b{-\f{1}{2} + \f{\sqrt{3}}{2}i} \qquad x_3 = \scbrt{a}\b{-\f{1}{2} - \f{\sqrt{3}}{2}i} \\[15pt] % The extra column sets the height of the row so that there will be more % vertical padding at the top and bottom % The rows are align middle. % The \!\!\!\!\!\!\!\!\! removes the horizontal space that is % introduced by the extra column \begin{array}{l|ll} \hline \begin{split} & \scbrt{1} = 1 \\ & \box{x_1 = 1} \qquad x_2 = -\f{1}{2} + \f{\sqrt{3}}{2}i \qquad x_3 = -\f{1}{2} - \f{\sqrt{3}}{2}i \end{split} & \begin{split} & \scbrt{-1} = -1 \\ & x_1 = -1 \qquad x_2 = \f{1}{2} - \f{\sqrt{3}}{2}i \qquad \box{x_3 = \f{1}{2} + \f{\sqrt{3}}{2}i} \end{split} & \!\!\!\!\!\!\!\!\! \begin{split} & \\[25pt] & \end{split} \\ \hline \begin{split} & \scbrt{i} = -i \\ & x_1 = -i \qquad \box{x_2 = \f{\sqrt{3}}{2} + \f{1}{2}i} \qquad x_3 = -\f{\sqrt{3}}{2} + \f{1}{2}i \end{split} & \begin{split} & \scbrt{-i} = i \\ & x_1 = i \qquad x_2 = -\f{\sqrt{3}}{2} - \f{1}{2}i \qquad \box{x_3 = \f{\sqrt{3}}{2} - \f{1}{2}i} \end{split} & \!\!\!\!\!\!\!\!\! \begin{split} & \\[25pt] & \end{split} \end{array} \]