Math Problem Solver

The solutions are given by the formula: \[x = 2^{\frac{2}{3}} \cos{\left(\frac{\pi k}{3} \right)} + 2^{\frac{2}{3}} \sin{\left(\frac{\pi k}{3} \right)} i\] where \(k\) is an integer in \(0 \le k < 6\).

There are \(2\) real solutions.

There are \(4\) solutions with nonzero imaginary part.

Real solutions: \[\begin{aligned}x &= 2^{\frac{2}{3}} \\&\approx 1.5874011\\x &= - 2^{\frac{2}{3}} \\&\approx -1.5874011\end{aligned}\]

Solutions with nonzero imaginary part: \[\begin{aligned}x &= \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2} \\&\approx 0.79370053 + 1.3747296 i\\x &= - \frac{2^{\frac{2}{3}}}{2} + \frac{2^{\frac{2}{3}} \sqrt{3} i}{2} \\&\approx -0.79370053 + 1.3747296 i\\x &= - \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2} \\&\approx -0.79370053 -1.3747296 i\\x &= \frac{2^{\frac{2}{3}}}{2} - \frac{2^{\frac{2}{3}} \sqrt{3} i}{2} \\&\approx 0.79370053 -1.3747296 i\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).