【数学I】例題1.2.6：対称式の値（One More）★★

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% 例題I1.2.6：対称式の値（One More）★★
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$x+\frac{1}{x}=\sqrt{6}$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) $x-\frac{1}{x}$

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% 例題I1.2.6の解答（One More）★★
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(1) $x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\cdot\frac{1}{x}=(\sqrt{6})^2-2=4$

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% 例題I1.2.6の別解
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$x+\frac{1}{x}=\sqrt{6}$の両辺を2乗すると，$x^2+2+\frac{1}{x^2}=6$

よって，$x^2+\frac{1}{x^2}=4$

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% 例題I1.2.6の解答（One More）★★
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(2)

\begin{align*}
x^3+\frac{1}{x^3}&=(x+\frac{1}{x})^3-3x\cdot\frac{1}{x}(x+\frac{1}{x})=(\sqrt{6})^3-3\cdot 1\cdot\sqrt{6}\\
&=6\sqrt{6}-3\sqrt{6}=3\sqrt{6}
\end{align*}

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% 例題I1.2.6の別解
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\begin{align*}
x^3+\frac{1}{x^3}&=(x+\frac{1}{x})\{x^2-x\cdot\frac{1}{x}+(\frac{1}{x})^2\}\\
&=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)\\
&=\sqrt{6}(4-1)=3\sqrt{6}
\end{align*}

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% 例題I1.2.6の解答（One More）★★
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(3) $(x-\frac{1}{x})^2=x^2-2x\cdot\frac{1}{x}+\frac{1}{x^2}=(x^2+\frac{1}{x^2})-2=4-2=2$

したがって，$(x-\frac{1}{x})^2=2$

よって，$x-\frac{1}{x}=\pm\sqrt{2}$

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% 問題I1.2.6：対称式の値（One More）★★
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$x-\frac{1}{x}=3$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) $x^6+\frac{1}{x^6}$

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% 問題I1.2.6の解答（One More）★★
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(1) $x^2+\frac{1}{x^2}=(x-\frac{1}{x})^2+2x\cdot\frac{1}{x}=3^2+2=11$

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% 問題I1.2.6の別解
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$x-\frac{1}{x}=3$の両辺を2乗すると，$x^2-2+\frac{1}{x^2}=9$

よって，$x^2+\frac{1}{x^2}=11$

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% 問題I1.2.6の解答（One More）★★
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(2) $(x+\frac{1}{x})^2=x^2+2x\cdot\frac{1}{x}+\frac{1}{x^2}=x^2+\frac{1}{x^2}+2=11+2=13$

よって，$x+\frac{1}{x}=\pm\sqrt{13}$

(3)

\begin{align*}
x^3+\frac{1}{x^3}&=(x+\frac{1}{x})\{x^2-x\cdot\frac{1}{x}+(\frac{1}{x})^2\}\\
&=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)\\
&=\pm\sqrt{13}(11-1)=\pm 10\sqrt{13}
\end{align*}

(4)

\begin{align*}
x^6+\frac{1}{x^6}&=(x^2+\frac{1}{x^2})^3-3x^2\cdot\frac{1}{x^2}(x^2+\frac{1}{x^2})\\
&=11^3-3\cdot 11=1298
\end{align*}

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% 問題I1.2.6の別解
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$x^6+\frac{1}{x^6}=(x^3+\frac{1}{x^3})^2-2=(\pm 10\sqrt{13})^2-2=1300-2=1298$

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% 例題I1.2.6の側注：対称式の値（One More）★★
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% 側注1
$x^2+y^2=(x+y)^2-2xy$

% 側注2
\begin{align*}
&x^3+y^3\\
&=(x+y)^3-3xy(x+y)
\end{align*}

% 側注3
\begin{align*}
&x^3+y^3\\
=&(x+y)(x^2-xy+y^2)
\end{align*}

% 側注4
(1)の$x^2+\frac{1}{x^2}=4$を用いるために，$(x-\frac{1}{x})^2$を考える．

% 側注5
$(x-y)^2=(x+y)^2-4xy$を用いて求めてもよい．

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% 問題I1.2.6の側注：対称式の値（One More）★★
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% 側注1
$x^2+y^2=(x-y)^2+2xy$

% 側注2
(1)の結果より，
\[
x^2+\frac{1}{x^2}=11
\]

% 側注3
\begin{align*}
&x^3+y^3\\
&=(x+y)(x^2-xy+y^2)
\end{align*}

% 側注4
$x^3+y^3=(x+y)^3-3xy(x+y)$を用いて求めてもよい．

% 側注5
\begin{align*}
&x^3+y^3\\
&=(x+y)^3-3xy(x+y)
\end{align*}

% 側注6
$x^2+y^2=(x+y)^2-2xy$

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% 例題I1.2.6の考え方・補足など：対称式の値（One More）★★
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% 考え方
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(1)，(2)$x^2+y^2=(x+y)^2-2xy,x^3+y^3=(x+y)^3-3xy(x+y)$を用いる．

(3) $(x-\frac{1}{x})^2$を用いる．

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% 注意
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$\frac{1}{x}=y$とおくと，$x+\frac{1}{x}=x+y,x^2+\frac{1}{x^2}=x^2+y^2,x^3+\frac{1}{x^3}=x^3+y^3$のように，$x$と$y$の対称式となる．なお，$x-\frac{1}{x}$は対称式ではないが，$(x-\frac{1}{x})^2$は対称式である．