【数学I】例題1.2.7：3文字の対称式の値（One More）★★★

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

% 解答（例題I1.2.7）
(1)$x+y+z=(-1+\sqrt{3})+(-1-\sqrt{3})+2=0$(2)$\begin{aligned} &xy+yz+zx\\ =&(-1+\sqrt{3})(-1-\sqrt{3})+(-1-\sqrt{3}) \cdot 2+2 \cdot (-1+\sqrt{3})\\ =&1-3-2-2\sqrt{3}-2+2\sqrt{3}=-6 \end{aligned}$(3)$xyz=(-1+\sqrt{3})(-1-\sqrt{3}) \cdot 2=(1-3) \cdot 2=-4$(4)$\begin{aligned} \frac{1}{x}+\frac{1}{y}+\frac{1}{z}&=\frac{yz}{x \cdot yz}+\frac{zx}{y \cdot zx}+\frac{xy}{z \cdot xy}=\frac{yz+zx+xy}{xyz}\\ &=\frac{-6}{-4}=\frac{3}{2} \end{aligned}$(5)$\begin{aligned} x^2+y^2+z^2&=(x+y+z)^2-2(xy+yz+zx)\\ &=0^2-2 \cdot (-6)=12 \end{aligned}$(6)$\begin{aligned} x^3+y^3+z^3&=(x^3+y^3+z^3-3xyz)+3xyz\\ &=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz\\ &=0+3 \cdot (-4)=-12 \end{aligned}$

% 問題I1.2.7
$x+y+z=3,x y+y z+z x=1,x y z=-2$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)$x^3+y^3+z^3$

% 解答I1.2.7
(1)$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{y z}{x \cdot y z}+\frac{z x}{y \cdot z x}+\frac{x y}{z \cdot x y}=\frac{yz+zx+xy}{xyz}=\frac{1}{-2}=-\frac{1}{2}$(2)$\begin{aligned} x^2+y^2+z^2 &=(x+y+z)^2-2(xy+yz+zx)\\ &=3^2-2 \cdot 1 \\ &=9-2=7 \end{aligned}$(3)$\begin{aligned} x^3+y^3+z^3 &=(x^3+y^3+z^3-3xyz)+3xyz\\ &=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)+3xyz \\ &=3 \cdot (7-1)+3 \cdot (-2)=12 \end{aligned}$