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

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

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


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