# Körpererweiterung/X^3-3X+1/Nullstelle/Zerfällt/Aufgabe/Lösung

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Dies beruht unter Verwendung von

${\displaystyle {}\alpha ^{3}=3\alpha -1\,}$

auf den Rechnungen

{\displaystyle {}{\begin{aligned}\beta ^{3}-3\beta +1&={\left(\alpha ^{2}-2\right)}^{3}-3{\left(\alpha ^{2}-2\right)}+1\\&=\alpha ^{6}-6\alpha ^{4}+12\alpha ^{2}-8-3\alpha ^{2}+6+1\\&=\alpha ^{6}-6\alpha ^{4}+9\alpha ^{2}-1\\&=9\alpha ^{2}-6\alpha +1-6\alpha {\left(3\alpha -1\right)}+9\alpha ^{2}-1\\&=0\end{aligned}}}

und

{\displaystyle {}{\begin{aligned}\gamma ^{3}-3\gamma +1&=-{\left(\alpha ^{2}+\alpha -2\right)}^{3}+3{\left(\alpha ^{2}+\alpha -2\right)}+1\\&=-\alpha ^{6}-3\alpha ^{5}+3\alpha ^{4}+11\alpha ^{3}-6\alpha ^{2}-12\alpha +8+3\alpha ^{2}+3\alpha -6+1\\&=-{\left(3\alpha -1\right)}^{2}-3\alpha ^{2}{\left(3\alpha -1\right)}+3\alpha {\left(3\alpha -1\right)}+11{\left(3\alpha -1\right)}-3\alpha ^{2}-9\alpha +3\\&=-9\alpha ^{2}+6\alpha -9{\left(3\alpha -1\right)}+3\alpha ^{2}+9\alpha ^{2}-3\alpha +33\alpha -11-3\alpha ^{2}-9\alpha +2\\&=-9\alpha ^{2}+6\alpha -27\alpha +9+3\alpha ^{2}+9\alpha ^{2}-3\alpha +33\alpha -11-3\alpha ^{2}-9\alpha +2\\&=0.\end{aligned}}}