find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$  so $z=\frac{-11}{8}$  however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you

For any $z = a + ib \in \mathbb{C}$,
$$
z = \rho\left(\cos{\varphi} + i\sin{\varphi}\right),
$$
where
$$
\rho = \sqrt{a^2 + b^2}, \text{ and }\varphi = \arctan\left(\frac{b}{a}\right).
$$
Then,
$$
\sqrt[n]{z} = \sqrt[n]{\rho}\left(\cos{\left(\frac{\varphi + 2\pi k}{n}\right)} + i\sin{\left(\frac{\varphi + 2\pi k}{n}\right)}\right), \quad k = 0, 1, \ldots, n-1.
$$
In our example,
$$
\frac{1}{64} = \frac{1}{64} + i0, \text{ i. e., }a = \frac{1}{64}\text{ and }b = 0 \Rightarrow
$$
$$
\frac{1}{64} = \frac{1}{64}\left(\cos(0) + i\sin(0)\right) \Rightarrow
$$
$$
\sqrt[3]{\frac{1}{64}} = \sqrt[3]{\frac{1}{64}}\left(\cos\left(\frac{2\pi k}{3}\right) + i\sin\left(\frac{2\pi k}{3}\right)\right), \quad k = 0, 1, 2.
$$
$$
\begin{array}{rl}
k = 0 \Rightarrow & z_1 = \frac{1}{4} \\
k = 1 \Rightarrow & z_2 = \frac{1}{4}\left(\cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -\frac{1 - i\sqrt{3}}{8}\\
k = 2 \Rightarrow & z_3 = \frac{1}{4}\left(\cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right)\right) = \frac{1}{4}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -\frac{1 + i\sqrt{3}}{8}
\end{array}
$$
