What is the sum of imaginary roots of this equation? What is the sum of imaginary roots of equation:
$$x^3+3x^2+3x+3$$
Here is my attempt:
$$x^3+3x^2+3x+3=(x^3+3x^2+3x+1)+2=(x+1)^3+2$$
Since $f(x')=0$ if x' is a root of $f(x)$
$$(x'+1)^3+2=0$$
$$(x'+1)^3=-2$$
$$x'=-1+\sqrt[3]{-2}$$
$$x'=-1+\sqrt[3]{2}\sqrt[3]{-1}$$........(i)
Again,$$-1=cos(\pi)+isin(\pi)=cos(-\pi)+isin(-\pi)$$
By Euler's formula,
$$-1=e^{i\pi}=e^{-i\pi}$$
$$\sqrt[3]{-1}=e^{i\frac{\pi}{3}}= e^{i\frac{-\pi}{3}}$$
Substuting in equation..(i)
$$x'=-1+\sqrt[3]{2}e^{i\frac{\pi}{3}},-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})},-1-\sqrt[3]{2}$$
Sum of imaginary roots is:
$$ -1+\sqrt[3]{2}e^{i\frac{\pi}{3}}+-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})}$$
$$=-1+\sqrt[3]{2}e^{i\frac{\pi}{3}}+-1+\sqrt[3]{2}e^{i(\frac{-\pi}{3})}$$
$$=-2+\sqrt[3]{2}(e^{i\frac{\pi}{3}}+e^{-i\frac{\pi}{3}})$$
$$=-2+\sqrt[3]{2}[[cos(\frac{\pi}{3})+isin(\frac{\pi}{3})]+cos(-\frac{\pi}{3})+isin(\frac{-\pi}{3})]=-2+\sqrt[3]{2}$$
It does not match with the given answer in the book.I am confused.
Any hint on my mistakes is appreciated.

So,  the values of  $x+1$ are $$\sqrt[3]2, \sqrt[3]2w,\sqrt[3]2w^2$$ where $w$ is a complex cube root of unity
So, the only real root of the given equation is $$-1+\sqrt[3]2$$
Now the sum of all roots is $$-\dfrac31$$
