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問題アーカイブ01
問題アーカイブ01次の等式を証明せよ。
\({\small (1)}~\)\(x^4+x^2+1=(x^2+x+1)(x^2-x+1)\)
\({\small (2)}~\)\((a^2-b^2)(x^2-y^2)=(ax+by)^2-(ay+bx)^2\)
\({\small (1)}~\)\(x^4+x^2+1=(x^2+x+1)(x^2-x+1)\)
\({\small (2)}~\)\((a^2-b^2)(x^2-y^2)=(ax+by)^2-(ay+bx)^2\)
数研出版｜数学Ⅱ[709] p.28 練習20
\({\small (1)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&(x^2+x+1)(x^2-x+1)
\\[3pt]~~~&=&\{(x^2+1)+x\}\{(x^2+1)-x\}
\\[3pt]~~~&=&(x^2+1)^2-x^2
\\[3pt]~~~&=&x^4+2x^2+1-x^2
\\[3pt]~~~&=&x^4+x^2+1\end{eqnarray}\)
したがって、\(x^4+x^2+1=(x^2+x+1)(x^2-x+1)\) [終]
\({\small (2)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(a^2-b^2)(x^2-y^2)
\\[3pt]~~~&=&a^2x^2-a^2y^2-b^2x^2+b^2y^2\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&(ax+by)^2-(ay+bx)^2
\\[3pt]~~~&=&a^2x^2+2abxy+b^2y^2-(a^2y^2+2abxy+b^2x^2)
\\[3pt]~~~&=&a^2x^2+2abxy+b^2y^2-a^2y^2-2abxy-b^2x^2
\\[3pt]~~~&=&a^2x^2+b^2y^2-a^2y^2-b^2x^2
\\[3pt]~~~&=&a^2x^2-a^2y^2-b^2x^2+b^2y^2\end{eqnarray}\)
\\[3pt]~~~&=&a^2x^2+2abxy+b^2y^2-(a^2y^2+2abxy+b^2x^2)
\\[3pt]~~~&=&a^2x^2+2abxy+b^2y^2-a^2y^2-2abxy-b^2x^2
\\[3pt]~~~&=&a^2x^2+b^2y^2-a^2y^2-b^2x^2
\\[3pt]~~~&=&a^2x^2-a^2y^2-b^2x^2+b^2y^2\end{eqnarray}\)
したがって、
\((a^2-b^2)(x^2-y^2)=(ax+by)^2-(ay+bx)^2\) [終]
問題アーカイブ02
問題アーカイブ02次の等式を証明せよ。
\({\small (1)}~\)\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\)
\({\small (2)}~\)\(x^3-\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)^3+3\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)\)
\({\small (1)}~\)\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\)
\({\small (2)}~\)\(x^3-\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)^3+3\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)\)
数研出版｜数学Ⅱ[709] p.38 問題 9
\({\small (1)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)
\\[5pt]~~~&=&x^3+3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+3x+\displaystyle\frac{\,3\,}{\,x\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
\\[5pt]~~~&=&x^3+3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+3x+\displaystyle\frac{\,3\,}{\,x\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
したがって、\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\) [終]
\({\small (2)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)^3+3\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)
\\[5pt]~~~&=&x^3-3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}-\displaystyle\frac{\,1\,}{\,x^3\,}+3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3-3x+\displaystyle\frac{\,3\,}{\,x\,}-\displaystyle\frac{\,1\,}{\,x^3\,}+3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3-\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
\\[5pt]~~~&=&x^3-3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}-\displaystyle\frac{\,1\,}{\,x^3\,}+3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3-3x+\displaystyle\frac{\,3\,}{\,x\,}-\displaystyle\frac{\,1\,}{\,x^3\,}+3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3-\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
したがって、\(x^3-\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)^3+3\left(x-\displaystyle\frac{\,1\,}{\,x\,}\right)\) [終]
問題アーカイブ03
問題アーカイブ03次の等式を証明せよ。
\({\small (1)}~\)\(a^3-b^3=(a-b)^3+3ab(a-b)\)
\({\small (2)}~\)\(a^2+ab+b^2=\left(a+\displaystyle\frac{\,b\,}{\,2\,}\right)^2+\displaystyle\frac{\,3\,}{\,4\,}b^2\)
\({\small (3)}~\)\((1+x)^3=1+x+x(1+x)+x(1+x)^2\)
\({\small (1)}~\)\(a^3-b^3=(a-b)^3+3ab(a-b)\)
\({\small (2)}~\)\(a^2+ab+b^2=\left(a+\displaystyle\frac{\,b\,}{\,2\,}\right)^2+\displaystyle\frac{\,3\,}{\,4\,}b^2\)
\({\small (3)}~\)\((1+x)^3=1+x+x(1+x)+x(1+x)^2\)
数研出版｜高等学校数学Ⅱ[710] p.27 練習23
数研出版｜新編数学Ⅱ[711] p.26 練習22
\({\small (1)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&(a-b)^3+3ab(a-b)
\\[3pt]~~~&=&a^3-3a^2b+3ab^2-b^3+3a^2b-3ab^2
\\[3pt]~~~&=&a^3-b^3\end{eqnarray}\)
したがって、\(a^3-b^3=(a-b)^3+3ab(a-b)\) [終]
\({\small (2)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\left(a+\displaystyle\frac{\,b\,}{\,2\,}\right)^2+\displaystyle\frac{\,3\,}{\,4\,}b^2
\\[5pt]~~~&=&a^2+2\cdot a\cdot\displaystyle\frac{\,b\,}{\,2\,}+\displaystyle\frac{\,b^2\,}{\,4\,}+\displaystyle\frac{\,3\,}{\,4\,}b^2
\\[5pt]~~~&=&a^2+ab+\displaystyle\frac{\,1\,}{\,4\,}b^2+\displaystyle\frac{\,3\,}{\,4\,}b^2
\\[5pt]~~~&=&a^2+ab+b^2\end{eqnarray}\)
したがって、\(a^2+ab+b^2=\left(a+\displaystyle\frac{\,b\,}{\,2\,}\right)^2+\displaystyle\frac{\,3\,}{\,4\,}b^2\) [終]
\({\small (3)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(1+x)^3
\\[3pt]~~~&=&1+3x+3x^2+x^3\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&1+x+x(1+x)+x(1+x)^2
\\[3pt]~~~&=&1+x+x+x^2+x(1+2x+x^2)
\\[3pt]~~~&=&1+2x+x^2+x+2x^2+x^3
\\[3pt]~~~&=&1+3x+3x^2+x^3\end{eqnarray}\)
したがって、\((1+x)^3=1+x+x(1+x)+x(1+x)^2\) [終]
問題アーカイブ04
問題アーカイブ04次の等式を証明せよ。
\({\small (1)}~\)\(x^2+\displaystyle\frac{\,1\,}{\,x^2\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2\)
\({\small (2)}~\)\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\)
\({\small (1)}~\)\(x^2+\displaystyle\frac{\,1\,}{\,x^2\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2\)
\({\small (2)}~\)\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\)
数研出版｜高等学校数学Ⅱ[710] p.36 問題 9
数研出版｜新編数学Ⅱ[711] p.36 章末問題A 5
\({\small (1)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2
\\[5pt]~~~&=&x^2+2\cdot x\cdot\displaystyle\frac{\,1\,}{\,x\,}+\displaystyle\frac{\,1\,}{\,x^2\,}-2
\\[5pt]~~~&=&x^2+2+\displaystyle\frac{\,1\,}{\,x^2\,}-2
\\[5pt]~~~&=&x^2+\displaystyle\frac{\,1\,}{\,x^2\,}\end{eqnarray}\)
したがって、\(x^2+\displaystyle\frac{\,1\,}{\,x^2\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2\) [終]
\({\small (2)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)
\\[5pt]~~~&=&x^3+3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+3x+\displaystyle\frac{\,3\,}{\,x\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
\\[5pt]~~~&=&x^3+3\cdot x^2\cdot\displaystyle\frac{\,1\,}{\,x\,}+3\cdot x\cdot\displaystyle\frac{\,1\,}{\,x^2\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+3x+\displaystyle\frac{\,3\,}{\,x\,}+\displaystyle\frac{\,1\,}{\,x^3\,}-3x-\displaystyle\frac{\,3\,}{\,x\,}
\\[5pt]~~~&=&x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\end{eqnarray}\)
したがって、\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\) [終]
問題アーカイブ05
※ 等式の証明ではないが、問題アーカイブ04の等式を用いた計算問題です。
問題アーカイブ05
\({\small (1)}~x+\displaystyle\frac{\,1\,}{\,x\,}\)
\({\small (2)}~x^2+\displaystyle\frac{\,1\,}{\,x^2\,}\)
\({\small (3)}~x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\)
\(x=\displaystyle\frac{\,1\,}{\,\sqrt{5}+2\,}\) のとき、次の式の値を求めよ。
\({\small (1)}~x+\displaystyle\frac{\,1\,}{\,x\,}\)
\({\small (2)}~x^2+\displaystyle\frac{\,1\,}{\,x^2\,}\)
\({\small (3)}~x^3+\displaystyle\frac{\,1\,}{\,x^3\,}\)
数研出版｜高等学校数学Ⅱ[710] p.37 章末問題A 1
\({\small (1)}~\)
\(\begin{eqnarray}~~~x&=&\displaystyle\frac{\,1\,}{\,\sqrt{5}+2\,}
\\[5pt]~~~&=&\displaystyle\frac{\,\sqrt{5}-2\,}{\,(\sqrt{5}+2)(\sqrt{5}-2)\,}
\\[5pt]~~~&=&\displaystyle\frac{\,\sqrt{5}-2\,}{\,5-4\,}
\\[5pt]~~~&=&\sqrt{5}-2\end{eqnarray}\)
よって、\(\displaystyle\frac{\,1\,}{\,x\,}=\sqrt{5}+2\) となるので、
\(\begin{eqnarray}~~~x+\displaystyle\frac{\,1\,}{\,x\,}&=&(\sqrt{5}-2)+(\sqrt{5}+2)
\\[5pt]~~~&=&2\sqrt{5}\end{eqnarray}\)
したがって、\(x+\displaystyle\frac{\,1\,}{\,x\,}=2\sqrt{5}\)
\({\small (2)}~\)\({\small (1)}\) より、\(x+\displaystyle\frac{\,1\,}{\,x\,}=2\sqrt{5}\)
等式 \(x^2+\displaystyle\frac{\,1\,}{\,x^2\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2\) を用いると、
\(\begin{eqnarray}~~~x^2+\displaystyle\frac{\,1\,}{\,x^2\,}&=&\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^2-2
\\[5pt]~~~&=&(2\sqrt{5})^2-2
\\[5pt]~~~&=&20-2
\\[5pt]~~~&=&18\end{eqnarray}\)
したがって、\(x^2+\displaystyle\frac{\,1\,}{\,x^2\,}=18\)
\({\small (3)}~\)\({\small (1)}\) より、\(x+\displaystyle\frac{\,1\,}{\,x\,}=2\sqrt{5}\)
等式 \(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)\) を用いると、
\(\begin{eqnarray}~~~x^3+\displaystyle\frac{\,1\,}{\,x^3\,}&=&\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)^3-3\left(x+\displaystyle\frac{\,1\,}{\,x\,}\right)
\\[5pt]~~~&=&(2\sqrt{5})^3-3\cdot 2\sqrt{5}
\\[5pt]~~~&=&8\cdot 5\sqrt{5}-6\sqrt{5}
\\[5pt]~~~&=&40\sqrt{5}-6\sqrt{5}
\\[5pt]~~~&=&34\sqrt{5}\end{eqnarray}\)
\\[5pt]~~~&=&(2\sqrt{5})^3-3\cdot 2\sqrt{5}
\\[5pt]~~~&=&8\cdot 5\sqrt{5}-6\sqrt{5}
\\[5pt]~~~&=&40\sqrt{5}-6\sqrt{5}
\\[5pt]~~~&=&34\sqrt{5}\end{eqnarray}\)
したがって、\(x^3+\displaystyle\frac{\,1\,}{\,x^3\,}=34\sqrt{5}\)
問題アーカイブ06
問題アーカイブ06次の等式を証明せよ。
\({\small (1)}~\)\((x^2-1)^2+3x(x-1)=(x^2+1)^2-x(x+3)\)
\({\small (2)}~\)\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\)
\({\small (1)}~\)\((x^2-1)^2+3x(x-1)=(x^2+1)^2-x(x+3)\)
\({\small (2)}~\)\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\)
東京書籍｜Advanced数学Ⅱ[701] p.49 問3
\({\small (1)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(x^2-1)^2+3x(x-1)
\\[3pt]~~~&=&x^4-2x^2+1+3x^2-3x
\\[3pt]~~~&=&x^4+x^2-3x+1\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&(x^2+1)^2-x(x+3)
\\[3pt]~~~&=&x^4+2x^2+1-x^2-3x
\\[3pt]~~~&=&x^4+x^2-3x+1\end{eqnarray}\)
したがって、\((x^2-1)^2+3x(x-1)=(x^2+1)^2-x(x+3)\) [終]
\({\small (2)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(a^2+b^2)(c^2+d^2)
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&(ac+bd)^2+(ad-bc)^2
\\[3pt]~~~&=&a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2
\\[3pt]~~~&=&a^2c^2+b^2d^2+a^2d^2+b^2c^2
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
\\[3pt]~~~&=&a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2
\\[3pt]~~~&=&a^2c^2+b^2d^2+a^2d^2+b^2c^2
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
したがって、\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\) [終]
問題アーカイブ07
問題アーカイブ07次の等式を証明せよ。
\({\small (1)}~\)\((2a+b)^2-(a+2b)^2=3(a^2-b^2)\)
\({\small (2)}~\)\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\)
\({\small (1)}~\)\((2a+b)^2-(a+2b)^2=3(a^2-b^2)\)
\({\small (2)}~\)\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\)
東京書籍｜Standard数学Ⅱ[702] p.53 問4
\({\small (1)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(2a+b)^2-(a+2b)^2
\\[3pt]~~~&=&4a^2+4ab+b^2-(a^2+4ab+4b^2)
\\[3pt]~~~&=&4a^2+4ab+b^2-a^2-4ab-4b^2
\\[3pt]~~~&=&3a^2-3b^2
\\[3pt]~~~&=&3(a^2-b^2)\end{eqnarray}\)
したがって、\((2a+b)^2-(a+2b)^2=3(a^2-b^2)\) [終]
\({\small (2)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(a^2+b^2)(c^2+d^2)
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&(ac+bd)^2+(ad-bc)^2
\\[3pt]~~~&=&a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2
\\[3pt]~~~&=&a^2c^2+b^2d^2+a^2d^2+b^2c^2
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
\\[3pt]~~~&=&a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2
\\[3pt]~~~&=&a^2c^2+b^2d^2+a^2d^2+b^2c^2
\\[3pt]~~~&=&a^2c^2+a^2d^2+b^2c^2+b^2d^2\end{eqnarray}\)
したがって、\((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\) [終]
問題アーカイブ08
問題アーカイブ08次の等式を証明せよ。
\({\small (1)}~\)\((ax-by)^2-(ay-bx)^2=(a^2-b^2)(x^2-y^2)\)
\({\small (2)}~\)\(a^2+b^2+c^2-ab-bc-ca=\displaystyle\frac{\,1\,}{\,2\,}\{(a-b)^2+(b-c)^2+(c-a)^2\}\)
\({\small (3)}~\)\((x+1)^3-(3x^2+1)=(x-1)^3+(3x^2+1)\)
\({\small (2)}~\)\(a^2+b^2+c^2-ab-bc-ca=\displaystyle\frac{\,1\,}{\,2\,}\{(a-b)^2+(b-c)^2+(c-a)^2\}\)
\({\small (3)}~\)\((x+1)^3-(3x^2+1)=(x-1)^3+(3x^2+1)\)
東京書籍｜Standard数学Ⅱ[702] p.62 Training 33
\({\small (1)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(ax-by)^2-(ay-bx)^2
\\[3pt]~~~&=&a^2x^2-2abxy+b^2y^2-(a^2y^2-2abxy+b^2x^2)
\\[3pt]~~~&=&a^2x^2-2abxy+b^2y^2-a^2y^2+2abxy-b^2x^2
\\[3pt]~~~&=&a^2x^2+b^2y^2-a^2y^2-b^2x^2
\\[3pt]~~~&=&a^2(x^2-y^2)-b^2(x^2-y^2)
\\[3pt]~~~&=&(a^2-b^2)(x^2-y^2)\end{eqnarray}\)
\\[3pt]~~~&=&a^2x^2-2abxy+b^2y^2-(a^2y^2-2abxy+b^2x^2)
\\[3pt]~~~&=&a^2x^2-2abxy+b^2y^2-a^2y^2+2abxy-b^2x^2
\\[3pt]~~~&=&a^2x^2+b^2y^2-a^2y^2-b^2x^2
\\[3pt]~~~&=&a^2(x^2-y^2)-b^2(x^2-y^2)
\\[3pt]~~~&=&(a^2-b^2)(x^2-y^2)\end{eqnarray}\)
したがって、\((ax-by)^2-(ay-bx)^2=(a^2-b^2)(x^2-y^2)\) [終]
\({\small (2)}~\)[証明]
(右辺)
\(\begin{eqnarray}~~~&=&\displaystyle\frac{\,1\,}{\,2\,}\{(a-b)^2+(b-c)^2+(c-a)^2\}
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,2\,}(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2)
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,2\,}(2a^2+2b^2+2c^2-2ab-2bc-2ca)
\\[5pt]~~~&=&a^2+b^2+c^2-ab-bc-ca\end{eqnarray}\)
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,2\,}(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2)
\\[5pt]~~~&=&\displaystyle\frac{\,1\,}{\,2\,}(2a^2+2b^2+2c^2-2ab-2bc-2ca)
\\[5pt]~~~&=&a^2+b^2+c^2-ab-bc-ca\end{eqnarray}\)
したがって、
\(a^2+b^2+c^2-ab-bc-ca=\displaystyle\frac{\,1\,}{\,2\,}\{(a-b)^2+(b-c)^2+(c-a)^2\}\) [終]
\({\small (3)}~\)[証明]
(左辺)
\(\begin{eqnarray}~~~&=&(x+1)^3-(3x^2+1)
\\[3pt]~~~&=&x^3+3x^2+3x+1-3x^2-1
\\[3pt]~~~&=&x^3+3x\end{eqnarray}\)
(右辺)
\(\begin{eqnarray}~~~&=&(x-1)^3+(3x^2+1)
\\[3pt]~~~&=&x^3-3x^2+3x-1+3x^2+1
\\[3pt]~~~&=&x^3+3x\end{eqnarray}\)
したがって、
\((x+1)^3-(3x^2+1)=(x-1)^3+(3x^2+1)\) [終]
