mateforum.ro

Sa se arate ca \( \sqrt[3]{3} \) nu se afla in \( \mathbb{Q}(\sqrt[3]{2}) \).

Presupunem ca \( \sqrt[3]{3}\in \mathbb{Q}(\sqrt[3]{2}) \). Atunci exista \( a,b,c \in \mathbb{Q} \) astfel incat \( 3=(a+b\sqrt[3]{2}+c\sqrt[3]{4})^3 \Rightarrow \left\{ \begin{array}{rcl} a^3+12abc+2b^3+4c^3=3 \\ a^2b+2ac^2+2b^2c=0 \\ a^2c+ab^2+2bc^2=0 \end{array} \Rightarrow a=0 \), fals sau \( b=c\sqrt[3]{2} \), fals fiindca \( a,b,c \in \mathbb{Q} \).