【数学I】例題1.2.5：対称式x^n+y^nの値（One More）★★★

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% 例題I1.2.5：対称式$x^n+y^n$の値（One More）★★★
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$x=\frac{2}{\sqrt{3}+\sqrt{2}},y=\frac{2}{\sqrt{3}-\sqrt{2}}$のとき，次の値を求めよ．

(1) $x+y$

(2) $xy$

(3) $x^2+y^2$

(4) $x^3+y^3$

(5) $x^4+y^4$

(6) $x^5+y^5$

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% 例題I1.2.5の解答（One More）★★★
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(1)

\begin{align*}
x+y&=\frac{2}{\sqrt{3}+\sqrt{2}}+\frac{2}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})+2(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\\
&=\frac{4\sqrt{3}}{3-2}=4\sqrt{3}
\end{align*}

(2) $xy=\frac{2}{\sqrt{3}+\sqrt{2}}\cdot\frac{2}{\sqrt{3}-\sqrt{2}}=\frac{4}{3-2}=4$

(3) $x^2+y^2=(x+y)^2-2xy=(4\sqrt{3})^2-2\cdot 4=48-8=40$

(4)

\begin{align*}
x^3+y^3&=(x+y)^3-3xy(x+y)=(4\sqrt{3})^3-3\cdot 4\cdot 4\sqrt{3}\\
&=192\sqrt{3}-48\sqrt{3}=144\sqrt{3}
\end{align*}

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% 例題I1.2.5の別解
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$x^3+y^3=(x+y)(x^2-xy+y^2)=4\sqrt{3}\cdot(40-4)=144\sqrt{3}$

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% 例題I1.2.5の解答（One More）★★★
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(5) $x^4+y^4=(x^2+y^2)^2-2x^2y^2=40^2-2\cdot 4^2=1600-32=1568$

(6)

\begin{align*}
x^5+y^5&=(x^2+y^2)(x^3+y^3)-x^2y^3-x^3y^2\\
&=(x^2+y^2)(x^3+y^3)-(xy)^2(x+y)\\
&=40\cdot 144\sqrt{3}-4^2\cdot 4\sqrt{3}=5760\sqrt{3}-64\sqrt{3}=5696\sqrt{3}
\end{align*}

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% 問題I1.2.5：対称式$x^n+y^n$の値（One More）★★★
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$x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}},y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$のとき，次の値を求めよ．

(1) $x+y$

(2) $xy$

(3) $x^2+y^2$

(4) $x^3+y^3$

(5) $x^4+y^4$

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% 問題I1.2.5の解答（One More）★★★
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(1)

\begin{align*}
x+y&=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{(\sqrt{5}-\sqrt{3})^2+(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\
&=\frac{(5-2\sqrt{15}+3)+(5+2\sqrt{15}+3)}{5-3}=8
\end{align*}

(2) $xy=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\cdot\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=1$

(3) $x^2+y^2=(x+y)^2-2xy=8^2-2\cdot 1=64-2=62$

(4) $x^3+y^3=(x+y)^3-3xy(x+y)=8^3-3\cdot 1\cdot 8=512-24=488$

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% 問題I1.2.5の別解
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$x^3+y^3=(x+y)(x^2-xy+y^2)=8\cdot(62-1)=488$

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% 問題I1.2.5の解答（One More）★★★
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(5) $x^4+y^4=(x^2+y^2)^2-2x^2y^2=62^2-2\cdot 1^2=3844-2=3842$

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% 例題I1.2.5の側注：対称式$x^n+y^n$の値（One More）★★★
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% 側注1
$x=\frac{2}{\sqrt{3}+\sqrt{2}}=2\sqrt{3}-2\sqrt{2},y=\frac{2}{\sqrt{3}-\sqrt{2}}=2\sqrt{3}+2\sqrt{2}$のように，$x,y$のそれぞれの分母を有理化してから，$x+y,xy$の値を計算してもよい．

% 側注2
(1)，(2)で求めた$x+y=4\sqrt{3},xy=4$を利用する．

% 側注3
\begin{align*}
&a^3+b^3\\
=&(a+b)(a^2-ab+b^2)
\end{align*}

% 側注4
\begin{align*}
&(x^2+y^2)^2\\
=&x^4+2x^2y^2+y^4
\end{align*}

% 側注5
\begin{align*}
&(x^2+y^2)(x^3+y^3)\\
=&x^5+x^2y^3+x^3y^2+y^5
\end{align*}

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% 問題I1.2.5の側注：対称式$x^n+y^n$の値（One More）★★★
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% 側注1
$x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=4-\sqrt{15},$

$y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=4+\sqrt{15}$のように，$x,y$のそれぞれの分母を有理化してから，$x+y,xy$の値を計算してもよい．

% 側注2
(1)，(2)で求めた$x+y=8,xy=1$を利用する．

% 側注3
\begin{align*}
&a^3+b^3\\
=&(a+b)(a^2-ab+b^2)
\end{align*}

% 側注4
\begin{align*}
&(x^2+y^2)^2\\
=&x^4+2x^2y^2+y^4
\end{align*}

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% 例題I1.2.5の考え方・補足など：対称式$x^n+y^n$の値（One More）★★★
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% 考え方
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$x+y,x^2+y^2$のように，$x,y$を入れ替えても，もとの式と同じ式になるものを，$x,y$の対称式といい，$x+y,xy$を基本対称式という．$x,y$の対称式は，基本対称式$x+y,xy$で表されることが知られている．

(3)，(4)$(x+y)^2=x^2+2xy+y^2$より，$x^2+y^2=(x+y)^2-2xy$と式変形できる．3次式の場合も同様に，$(x+y)^3=x^3+3x^2y+3xy^2+y^3$より，$x^3+y^3=(x+y)^3-3xy(x+y)$と式変形できることを用いればよい．

(6) $x^5=x^2\times x^3,y^5=y^2\times y^3$より，

\[
(x^2+y^2)(x^3+y^3)=x^5+x^2y^3+x^3y^2+y^5=x^5+y^5+x^2y^2(x+y)=x^5+y^5+(xy)^2(x+y)
\]

よって，$x^5+y^5=(x^2+y^2)(x^3+y^3)-(xy)^2(x+y)$となり，(1)〜(4)の結果を利用すればよい．

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$x,y$の対称式は，基本対称式$x+y,xy$で表す．

\[
x^2+y^2=(x+y)^2-2xy,x^3+y^3=(x+y)^3-3xy(x+y)
\]