Mahesh Godavarti
1

I think you actually mean x^3 + 3x^2 + 3x + 3 = 0 which can be rewritten as (x + 1)^3 = -2 .

APPROACH 1:

That means $x + 1 = -\sqrt[3]{2}, -\sqrt[3]{2} \omega, -\sqrt[3]{2} \omega^2$ where \omega is the cube root of unity.

Therefore, the non-real roots of x^3 + 3x^2 + 3x + 3 = 0 are $-\sqrt[3]{2} \omega - 1$ and $-\sqrt[3]{2} \omega^2 - 1$.

The sum of non-real roots is $-2 -\sqrt[3]{2}(\omega + \omega^2) = -2 - \sqrt[3]{2} \cdot -1 = -2 + \sqrt[3]{2}$.

This follows from the fact that 1 + \omega + \omega^2 = 0 .

APPROACH 2:

Use the fact that the sum of all roots of a polynomial a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 is -\frac{a_{n-1}}{a_n} .

Therefore, the sum of all roots of x^3 + 3x^2 + 3x + 3 = 0 is -3.

Now, we can simply find the sum of all non-real roots by subtracting the real root from the sum of all roots.

Therefore, the sum we want is $-3-(-1-\sqrt[3]{2})=-2+\sqrt[3]{2}$.