【数学Ⅰ】対称式の基本と因数分解など全問題

\[ \small{ \begin{align}
x+y & = \frac{\sqrt{3}+1}{\sqrt{3}-1} + \frac{\sqrt{3}-1}{\sqrt{3}+1} \\
\\
& = \frac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)} （←通分） \\
\\
& = \frac{(3+2\sqrt{3}+1)+(3-2\sqrt{3}+1)}{3-1} \\
\\
& = \frac{8}{2} \\
\\
& = 4
\end{align} } \]

\[ \begin{align}
& (x^2+y^2)(x^3+y^3) \\
\\
= & x^5+x^2y^3+x^3y^2+y^5 
\end{align} \]

より、

\[ \begin{align}
& x^5+y^5 \\
\\
= & (x^2+y^2)(x^3+y^3)-x^2y^3-x^3y^2 \\
\\
= & (x^2+y^2)(x^3+y^3)-(x+y)(xy)^2
\end{align} \]

(1)～(3)の結果より，

\[ \begin{align}
x^5+y^5 & = 14 \cdot 52-4 \cdot 1^2 \\
\\
& = 728-4 \\
\\
& = 724
\end{align} \]

\( (x+y)(x^4+y^4) = x^5+xy^4+x^4y+y^5 \)より、

\[ \begin{align}
& x^5+y^5 \\
\\
= & (x+y)(x^4+y^4)-xy^4-x^4y \\
\\
= & (x+y)(x^4+y^4)-xy(x^3+y^3) \\
\end{align} \]

(1)，(3)，(4)の結果より、

\[ \begin{align}
x^5+y^5 & = 4 \cdot 194-1 \cdot 52 \\
\\
& = 776-52 \\
\\
& = 724
\end{align} \]

\( x^4+y^4 = (x^2+y^2)^2-2x^2y^2 \)より、

\[ \begin{align}
x^4+\frac{1}{x^4} & = \left( x^2+\frac{1}{x^2} \right)^2-2x^2 \cdot \frac{1}{x^2} 
\end{align} \]

(1)の結果より，

\[ \begin{align}
x^4+\frac{1}{x^4} & = 5^2-2 \\
\\
& = 25-2 \\
\\
& = 23
\end{align} \]

\[ \begin{align}
& x^5+y^5 \\
\\
= & (x^2+y^2)(x^3+y^3)-x^2y^3-x^3y^2
\end{align} \]

より、

\[ \small{ \begin{align}
& x^5+y^5 \\
\\
= & \left( x^2+\frac{1}{x^2} \right) \left( x^3+\frac{1}{x^3} \right) – x^2 \cdot \frac{1}{x^3} – x^3 \cdot \frac{1}{x^2} \\
\\
= & \left( x^2+\frac{1}{x^2} \right) \left( x^3+\frac{1}{x^3} \right) – \left( x+\frac{1}{x} \right)
\end{align} } \]

(1)，(2)の結果より，

\[ \begin{align}
x^5+y^5 & = 5 \cdot 4\sqrt{7} – \sqrt{7} \\
\\
& = 20\sqrt{7}-\sqrt{7} \\
\\
& = 19\sqrt{7}
\end{align} \]

\( x^5+y^5 = (x+y)(x^4+y^4)-xy^4-x^4y \)より、

\[ \small{ \begin{align}
& x^5+y^5 \\
\\
= & \left( x+\frac{1}{x} \right) \left( x^4 + \frac{1}{x^4} \right) – x \cdot \frac{1}{x^4} – x^4 \cdot \frac{1}{x} \\
\\
= & \left( x+\frac{1}{x} \right) \left( x^4 + \frac{1}{x^4} \right) – \left( x^3 + \frac{1}{x^3} \right)
\end{align} } \]

(2)，(3)の結果より，

\[ \begin{align}
x^5+y^5 & = \sqrt{7} \cdot 23 – 4\sqrt{7} \\
\\
& = 23\sqrt{7} – 4\sqrt{7} \\
\\
& = 19\sqrt{7}
\end{align} \]

\[ \begin{align}
& x^3+y^3+z^3 \\
\\
= & (x+y+z)(x^2+y^2+z^2-xy-yz-zx) +3xyz
\end{align} \]

が成り立つから，(1)の結果より，

\[ \begin{align}
x^3+y^3+z^3 & = 1 \cdot (-3-2)+3 \cdot 3 \\
\\
& = -5+9 \\
\\
& = 4
\end{align} \]

\[ \small{ \begin{align}
& \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \\
\\
= & \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 -2 \left( \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} \right) \\
\\
= & \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 -2 \left( \frac{z}{xy \cdot z}+\frac{x}{yz \cdot x}+\frac{y}{zx \cdot y} \right) \\
\\
= & \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)^2 -2 \left( \frac{z+x+y}{xyz} \right)
\end{align} } \]

(3)の結果より，

\[ \begin{align}
\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} & = \left( \frac{2}{3} \right)^2 -2 \cdot \frac{1}{3} \\
\\
& = \frac{4}{9}-\frac{2}{3} \\
\\
& = – \frac{2}{9}
\end{align} \]