# sum of coefficients of $k(x)$

The polynomial $$x^3-3x^2-4x+4=0$$ has $$3$$ real roots $$\alpha,\beta,\gamma$$ and equation $$k(x)=x^3+ax^2+bx+c=0$$ has $$3$$ roots $$\alpha',\beta',\gamma'$$ and $$\alpha'=\alpha+\beta\omega+\gamma \omega^2\;$$ and $$\,\beta'=\alpha\omega+\beta\omega^2+\gamma\;,$$ $$\gamma'=\alpha\omega^2+\beta+\gamma\omega.$$ where $$\omega,\omega^2$$ are complex cube root of unity. Then absolute value of real parts of sum of coefficients of $$k(x)$$ is

what I tried $$\sum \alpha=3,\sum \alpha \beta=-4\;,\alpha \beta \gamma=-4$$

and $$\sum \alpha'=-a,\sum \alpha' \beta'=b\;,\alpha'\beta'\gamma'=-c$$

and $$\sum \alpha' =0$$

$$\sum \alpha'\beta'=(\sum\alpha^2 )\omega+2\sum \alpha \beta+2\omega(\sum \alpha \beta)+2\omega^2(\sum \alpha \beta)+\omega^2(\sum \alpha^2)+\sum \alpha^2=0$$

and $$\alpha'\beta'\gamma'=\sum \alpha^3+2\alpha \beta \gamma$$

How do I solve it? Help me please

• $$\beta'=\alpha'\omega, \gamma'=\alpha'\omega^2$$ – lab bhattacharjee Mar 16 at 15:51

## 1 Answer

Now, use $$\alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(\alpha+\beta+\gamma)(\alpha\beta+\alpha\gamma+\beta\gamma)+3\alpha\beta\gamma.$$