Find the zeros of $f(z)=z^3-\sin^3z$ I want to find the zeros of $f(z)$, $$f(z)=z^3-\sin^3z$$
My attempt
$f(z)=0$
$z^3-(z-z^3/3!+z^5/5!-\dots)^3=0$
$z^3-z^3(1-z^2/3!+z^4/5!-\dots)^3=0$
$z^3[1-(1-z^2/3!+z^4/5!-\dots)^3]=0$
So $z=0$ is a zero of order $3$. 
I don't feel good about this answer. Please give me some hints if I am incorrect. 
Edit: What is the order of root $z=0$?
 A: Interesting problem. I'm considering a couple of approaches. 
We have $f(z)=z^3-sin^3(z)$ and wish to find the roots, or equivalently solutions of $z^3=sin^3(z)$ which gives $$z^3/sin^3(z)=1$$
So roots occur where $z/sin(z)$= cubic roots of unity:$\{1, e^{2\pi i/3}, e^{4\pi i/3}\}$. 
Or consider $z^3-sin^3{z}=0=(z-\sin{z})(z^2+z\sin{z}+\sin^2z)$
Applying the quadratic formula to the second term in parentheses gives $$z=\frac{-\sin{z}+\sqrt{\sin^2{z}-4\sin^2{z}}}{2}=(\sin z)(\frac{-1}{2}+\frac{i\sqrt{3}}{2})$$
Again we get that $\frac{z}{\sin{z}}$=1 or the other two roots of unity. 
For each root of unity, use your Taylor Series representation of Sine to get the Taylor series of $\frac{\sin{z}}{z}$ to the quadratic term, set this equal to the successive roots of unity. This should get you close approximations to  several solutions.
A: Hint:  The root at zero is more than fourth order.  Also notice that 
$z^3 - \sin^3 z \approx 0$ 
when  $z= 5.3341 - 2.4614 i$.  That should lead you to some more roots.
You can compute the lower order terms of the Taylor Series from the algebra in your "My Attempt" section.
$$z^3 - \sin^3z = z^3 - (z-z^3/3!+ z^5/5!-z^7/7!+z^9/9!-O\left(z^{11}\right))^3$$
$$= z^3 - \left(z^3-\frac{z^5}{2}+\frac{13 z^7}{120}-\frac{41 z^9}{3024}+O\left(z^{11}\right)\right)$$
$$=\frac{z^5}{2}-\frac{13 z^7}{120}+\frac{41 z^9}{3024}-O\left(z^{11}\right)$$
So, fifth order at $z=0$.
