t^-3=64
You asked:
Solve the equation \({t}^{-3} = 64\) for the variable \(t\).
MathBot Answer:
The solutions are given by the formula: \[t = \frac{\cos{\left(\frac{2 \pi k}{3} \right)}}{4} + \frac{\sin{\left(\frac{2 \pi k}{3} \right)}}{4} i\] where \(k\) is an integer in \(0 \le k < 3\).
There is \(1\) real solution.There are \(2\) solutions with nonzero imaginary part.
Real solutions: \[\begin{aligned}t &= \frac{1}{4} \\&= 0.25\end{aligned}\]
Solutions with nonzero imaginary part: \[\begin{aligned}t &= - \frac{1}{8} + \frac{\sqrt{3} i}{8} \\&\approx -0.125 + 0.21650635 i\\t &= - \frac{1}{8} - \frac{\sqrt{3} i}{8} \\&\approx -0.125 -0.21650635 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).