What is the sum of imaginary roots of this equation?

Question

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.

If you mean the non-real roots I get $\sqrt[3]{2}-2$ too. What is the answer in the book? — gammatester, Oct 19 '16 at 10:10
It does not match, may be but this is the correct solution. You can doublecheck using Cardano's method. — Claude Leibovici, Oct 19 '16 at 10:10

score 2 · Answer 1 · answered Oct 19 '16 at 10:10

2

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 $$-\dfrac31$$

answered Oct 19 '16 at 10:10

lab bhattacharjee

1 Answers1