곱셈공식 수 계산

[출제의도] 인수분해를 이용하여 복잡한 계산을 간단히 한다. $218$을 $n$이라 하면 $218 ^{3} + 1$$= n ^{3} + 1$$= \left( n + 1 \right) \left( n ^{2} - n + 1 \right)$ $217 ^{3} - 1$$= \left( n - 1 \right) ^{3} - 1$ $= \left\{ \left( n - 1 \right) - 1 \right\} \left\{ \left( n - 1 \right) ^{2} + \left( n - 1 \right) + 1 \right\}$ $= \left( n - 2 \right) \left( n ^{2} - n + 1 \right)$ $\displaystyle \frac{218 ^{3} + 1}{217 ^{3} - 1}$$\displaystyle = \frac{\left( n + 1 \right) \left( n ^{2} - n + 1 \right)}{\left( n - 2 \right) \left( n ^{2} - n + 1 \right)}$ $\displaystyle = \frac{n + 1}{n - 2}$$\displaystyle = \frac{218 + 1}{218 - 2}$$\displaystyle = \frac{219}{216}$$\displaystyle = \frac{73}{72}$ [다른 풀이] $217$을 $n$이라 하면 $218 ^{3} + 1$$= \left( n + 1 \right) ^{3} + 1$ $= \left\{ \left( n + 1 \right) + 1 \right\} \left\{ \left( n + 1 \right) ^{2} - \left( n + 1 \right) + 1 \right\}$ $= \left( n + 2 \right) \left( n ^{2} + n + 1 \right)$ $217 ^{3} - 1$$= n ^{3} - 1$$= \left( n - 1 \right) \left( n ^{2} + n + 1 \right)$ $\displaystyle \frac{218 ^{3} + 1}{217 ^{3} - 1}$$\displaystyle = \frac{\left( n + 2 \right) \left( n ^{2} + n + 1 \right)}{\left( n - 1 \right) \left( n ^{2} + n + 1 \right)}$ $\displaystyle = \frac{n + 2}{n - 1}$$\displaystyle = \frac{217 + 2}{217 - 1}$$\displaystyle = \frac{219}{216}$$\displaystyle = \frac{73}{72}$