무리수 상등 빈칸

[출제의도] 고차방정식의 해 추론하기 $\displaystyle \sqrt{n} = a + b$($a$는 자연수, $0 < b < 1$)에서 $\displaystyle b = \sqrt{n} - a$ $\displaystyle a ^{3} - 9 ab + b ^{3} = a ^{3} - 9 a \left( \sqrt{n} - a \right) + \left( \sqrt{n} - a \right) ^{3}$ $\displaystyle = a ^{3} - 9 a \left( \sqrt{n} - a \right) + \left( n \sqrt{n} - 3 an + 3 a ^{2} \sqrt{n} - a ^{3} \right)$ $\displaystyle = \left( 3 a ^{2} - 9 a + n \right) \sqrt{n} + 9 a ^{2} + \left( {\square {- 3 n}} \right) \times a$ $= 0$ $\therefore$$\left( \text{가} \right) = - 3 n$ 무리수가 서로 같을 조건에 의하여 $3 a ^{2} - 9 a + n = 0$ $\cdots \cdots \text{㉠}$ $9 a ^{2} + \left( {\square {- 3 n}} \right) \times a = 0 \cdots \cdots \text{㉡}$ $\text{㉠}$, $\text{㉡}$에서 $\text{㉠} \times 3 a + \text{㉡}$에 의하여 $n$을 소거하면 $9 a ^{3} + {\square {- 18 a ^{2}}} = 0$ $\therefore$$\left( \text{나} \right) = - 18 a ^{2}$ $\therefore$$f \left( n \right) = - 3 n$, $g \left( a \right) = - 18 a ^{2}$ 따라서 $\displaystyle \frac{g \left( 2 \right)}{f \left( 2 \right)} = 12$