<s>[INST] 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
 [/INST] There is a simpler method. Just use
$$\begin{align}a^3=b^3 \Longleftrightarrow a^3-b^3=0\Longleftrightarrow (a-b)(a^2+ab+b^2)=0\end{align}$$
To find the complex roots, just  solve the quadratic:
$$a^2+ab+b^2=0.$$

In your case you can take $$a=2z+3, b=\dfrac 14.$$
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