【数学I】例題1.2.7：3文字の対称式の値（One More）★★★

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% 例題I1.2.7：3文字の対称式の値（One More）★★★
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$x=-1+\sqrt{3},y=-1-\sqrt{3},z=2$のとき，次の値を求めよ．

(1) $x+y+z$

(2) $xy+yz+zx$

(3) $xyz$

(4) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

(5) $x^2+y^2+z^2$

(6) $x^3+y^3+z^3$

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% 例題I1.2.7の解答（One More）★★★
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(1) $x+y+z=(-1+\sqrt{3})+(-1-\sqrt{3})+2=0$

(2)

\begin{align*}
&xy+yz+zx\\
=&(-1+\sqrt{3})(-1-\sqrt{3})+(-1-\sqrt{3})\cdot 2+2\cdot(-1+\sqrt{3})\\
=&1-3-2-2\sqrt{3}-2+2\sqrt{3}=-6
\end{align*}

(3) $xyz=(-1+\sqrt{3})(-1-\sqrt{3})\cdot 2=(1-3)\cdot 2=-4$

(4)

\begin{align*}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}&=\frac{yz}{x\cdot yz}+\frac{zx}{y\cdot zx}+\frac{xy}{z\cdot xy}=\frac{yz+zx+xy}{xyz}\\
&=\frac{-6}{-4}=\frac{3}{2}
\end{align*}

(5)

\begin{align*}
x^2+y^2+z^2&=(x+y+z)^2-2(xy+yz+zx)\\
&=0^2-2\cdot(-6)=12
\end{align*}

(6)

\begin{align*}
x^3+y^3+z^3&=(x^3+y^3+z^3-3xyz)+3xyz\\
&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz\\
&=0+3\cdot(-4)=-12
\end{align*}

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% 問題I1.2.7：3文字の対称式の値（One More）★★★
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$x+y+z=3,xy+yz+zx=1,xyz=-2$を満たす実数$x,y,z$に対して，次の式の値を求めよ．

(1) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$

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

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

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% 問題I1.2.7の解答（One More）★★★
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(1) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz}{x\cdot yz}+\frac{zx}{y\cdot zx}+\frac{xy}{z\cdot xy}=\frac{yz+zx+xy}{xyz}=\frac{1}{-2}=-\frac{1}{2}$

(2)

\begin{align*}
x^2+y^2+z^2&=(x+y+z)^2-2(xy+yz+zx)\\
&=3^2-2\cdot 1\\
&=9-2=7
\end{align*}

(3)

\begin{align*}
x^3+y^3+z^3&=(x^3+y^3+z^3-3xyz)+3xyz\\
&=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz\\
&=3\cdot(7-1)+3\cdot(-2)=12
\end{align*}

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% 例題I1.2.7の側注：3文字の対称式の値（One More）★★★
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% 側注1
$(a+b)(a-b)=a^2-b^2$

% 側注2
通分して分母を揃える．

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

% 側注4
(i)を利用する．

% 側注5
(3)より，$xyz=-4$であり，(1)より，$x+y+z=0$である．

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% 問題I1.2.7の側注：3文字の対称式の値（One More）★★★
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% 側注1
\begin{align*}
&(x+y+z)^2\\
=&x^2+y^2+z^2\\
&+2(xy+yz+zx)
\end{align*}

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% 例題I1.2.7の考え方・補足など：3文字の対称式の値（One More）★★★
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% 考え方
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$x,y,z$の対称式を基本対称式で表すことを考える．3文字の場合の基本対称式は，$x+y+z,xy+yz+zx,xyz$である．対称式は，基本対称式で表すことができる．

(5) 直接代入するのではなく，乗法公式$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$の利用を考える．$2xy+2yz+2zx$を移項した，次の式を用いる．

\[
x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)
\]

(6) $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\cdots(\mathrm{i})$を利用する．$-3abc$を移項した，次の式を利用すると考えてもよい．

\[
a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc
\]

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% One Point
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$x,y,z$の対称式は，基本対称式$x+y+z,xy+yz+zx,xyz$で表す．

\[
x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)
\]

\[
x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz
\]