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

Hope you got it now, it's a property of the polynomial, the equation $z^2-1$ is satisfied by $z =1, z= -1$, because a quadratic has two roots
While $z^3-1$ is satisfied by $z=1$ as the obvious real solution, but a cubic has three roots, so the other two are unreal, but if we can factor it or substitute $z=a+bi$, to get the remaining roots
$$ z^2+z+1$$
$$ z_1 = -\frac{1}{2}  + \frac{\sqrt{ -3}}{2}$$
$$ z_2 = -\frac{1}{2}  - \frac{\sqrt{ -3}}{2}$$
They are conjugate of eachother, If you cube it using the property of complex number the result is $z^3 = 1$, so $z_0 , z_1, z_2$ all from the cube root of unity
Anytime you take the cube root of a number $n$, the solutions are $\sqrt[3]{n} , \sqrt[3]{n} \cdot z_1, \sqrt[3]{n} \cdot z_2$
