삼차방정식 근과 계수

[출제의도] 삼차방정식을 이용하여 수학내적문제 해결하기 $x ^{3} + ax ^{2} + bx + c = 0$의 세 근이 $\alpha$, $\beta$, $\gamma$이므로 $\alpha + \beta + \gamma = - a$, $\alpha \beta + \beta \gamma + \gamma \alpha = b$, $\alpha \beta \gamma = - c$ $x ^{3} - 2 x ^{2} + 3 x - 1 = 0$의 세 근이 $\displaystyle \frac{1}{\alpha \beta}$, $\displaystyle \frac{1}{\beta \gamma}$, $\displaystyle \frac{1}{\gamma \alpha}$이므로 ⅰ) $\displaystyle \frac{1}{\alpha \beta} + \frac{1}{\beta \gamma} + \frac{1}{\gamma \alpha} = \frac{\alpha + \beta + \gamma}{\alpha \beta \gamma} = \frac{a}{c} = 2$ ⅱ) $\displaystyle \frac{1}{\alpha \beta} \times \frac{1}{\beta \gamma} + \frac{1}{\beta \gamma} \times \frac{1}{\gamma \alpha} + \frac{1}{\gamma \alpha} \times \frac{1}{\alpha \beta}$ $\displaystyle = \frac{1}{\alpha \beta ^{2} \gamma} + \frac{1}{\alpha \beta \gamma ^{2}} + \frac{1}{\alpha ^{2} \beta \gamma}$ $\displaystyle = \frac{\alpha \gamma + \alpha \beta + \beta \gamma}{( \alpha \beta \gamma ) ^{2}} = \frac{b}{c ^{2}} = 3$ ⅲ) $\displaystyle \frac{1}{\alpha \beta} \times \frac{1}{\beta \gamma} \times \frac{1}{\gamma \alpha} = \frac{1}{( \alpha \beta \gamma ) ^{2}} = \frac{1}{c ^{2}} = 1$ $c ^{2} = 1$이므로 ⅰ)에서 $a ^{2} = 4$, ⅱ)에서 $b ^{2} = 9$ $\therefore a ^{2} + b ^{2} + c ^{2} = 14$