【数学I】例題1.2.6：対称式の値（One More）★★

% 例題I1.2.6：対称式の値 （One More）★★
$x+\frac{1}{x}=\sqrt{6}$のとき,次の式の値を求めよ. (1)$x^2+\frac{1}{x^2}$(2)$x^3+\frac{1}{x^3}$(3)$x-\frac{1}{x}$

% 解答（例題I1.2.6）
(1)$x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2x \cdot \frac{1}{x}=(\sqrt{6})^2-2=4$別解：$x+\frac{1}{x}=\sqrt{6}$の両辺を2乗すると,$x^2+2+\frac{1}{x^2}=6$よって,$x^2+\frac{1}{x^2}=4$(2)$\begin{aligned} x^3+\frac{1}{x^3}&=\left(x+\frac{1}{x}\right)^3-3x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right)=(\sqrt{6})^3-3 \cdot 1 \cdot \sqrt{6}\\ &=6\sqrt{6}-3\sqrt{6}=3\sqrt{6} \end{aligned}$別解：$\begin{aligned} x^3+\frac{1}{x^3}&=\left(x+\frac{1}{x}\right)\left\{x^2-x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^2\right\}\\ &=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)\\ &=\sqrt{6}(4-1)=3\sqrt{6} \end{aligned}$(3)$\left(x-\frac{1}{x}\right)^2=x^2-2x \cdot \frac{1}{x}+\frac{1}{x^2}=\left(x^2+\frac{1}{x^2}\right)-2=4-2=2$したがって,$\left(x-\frac{1}{x}\right)^2=2$よって,$x-\frac{1}{x}=\pm\sqrt{2}$

% 問題I1.2.6
$x-\frac{1}{x}=3$のとき,次の式の値を求めよ. (1)$x^2+\frac{1}{x^2}$(2)$x+\frac{1}{x}$(3)$x^3+\frac{1}{x^3}$(4)$x^6+\frac{1}{x^6}$

% 解答I1.2.6
(1)$x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2x \cdot \frac{1}{x}=3^2+2=11$別解：$x-\frac{1}{x}=3$の両辺を2乗すると,$x^2-2+\frac{1}{x^2}=9$よって,$x^2+\frac{1}{x^2}=11$(2)$\left(x+\frac{1}{x}\right)^2 =x^2+2 x \cdot \frac{1}{x}+\frac{1}{x^2} =x^2+\frac{1}{x^2}+2=11+2=13$よって,$x+\frac{1}{x}=\pm\sqrt{13}$(3)$\begin{aligned} x^3+\frac{1}{x^3}&=\left(x+\frac{1}{x}\right)\left\{x^2-x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^2\right\}\\ &=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}-1\right)\\ &=\pm\sqrt{13}(11-1)=\pm10\sqrt{13} \end{aligned}$(4)$\begin{aligned} x^6+\frac{1}{x^6} &=\left(x^2+\frac{1}{x^2}\right)^3-3 x^2 \cdot \frac{1}{x^2}\left(x^2+\frac{1}{x^2}\right)\\ &=11^3-3 \cdot 11=1298 \end{aligned}$別解：$x^6+\frac{1}{x^6} =\left(x^3+\frac{1}{x^3}\right)^2-2 =(\pm 10\sqrt{13})^2-2 =1300-2=1298$