btakeshi
中学や数学１で扱った (a+b)^2 の展開公式を，３乗で考えてみましょう。何か法則に気がつくでしょうか。気づいた人は４乗・５乗と調べてみましょう。数学の美しさを感じてください。
和の２乗の展開公式
\large(a+b)^3 = a^3+3a^2b+3ab^2+b^3
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (a+b)^3\\
& \colMM{red}{\Darr ３乗をばらす}\\
&= (a+b)\colUL{red}{(a+b)(a+b)}\\
& \colMM{red}{\Darr 一部を展開}\\
&= (\colBX{bisque}{$a$}\,\colBX{palegreen}{$+b$})\colUL{red}{(a^2+2ab+b^2)}\\
& \colMM{red}{\Darr さらに展開 \Darr}\\
&= \colBX{bisque}{$a$}(a^2+2ab+b^2)\colBX{palegreen}{$+b$}(a^2+2ab+b^2)\\
&\\
&= a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\\
& \colMM{red}{\Darr 同類項を整理}\\
&= a^3+3a^2b+3ab^2+b^3 \colMM{red}{\cdots 完成！}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan\def\Left{a}
\def\Right{b}
\def\A{a^3}
\def\B{3a^2b}
\def\C{3ab^2}
\def\D{b^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan次の式を展開しなさい。
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{1}
\def\A{x^3}
\def\B{3x^2}
\def\C{3x}
\def\D{1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{2x}
\def\Right{y}
\def\A{8x^3}
\def\B{12x^2y}
\def\C{6xy^2}
\def\D{y^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{2}
\def\A{x^3}
\def\B{6x^2}
\def\C{12x}
\def\D{8}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{3a}
\def\Right{b}
\def\A{27a^3}
\def\B{27a^2b}
\def\C{9ab^2}
\def\D{b^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{2y}
\def\A{x^3}
\def\B{6x^2y}
\def\C{12xy^2}
\def\D{8y^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ +\ 右}\\
& (\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2+(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A + \B + \C + \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan差の２乗の展開公式
\large(a-b)^3=a^3-3a^2b+3ab^2-b^3
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (a-b)^3\\
& \colMM{red}{\Darr ３乗をばらす}\\
&= (a-b)\colUL{red}{(a-b)(a-b)}\\
& \colMM{red}{\Darr 一部を展開}\\
&= (\colBX{bisque}{$a$}\,\colBX{palegreen}{$-b$})\colUL{red}{(a^2-2ab+b^2)}\\
& \colMM{red}{\Darr さらに展開 \Darr}\\
&= \colBX{bisque}{$a$}(a^2-2ab+b^2)\colBX{palegreen}{$-b$}(a^2-2ab+b^2)\\
&\\
&= a^3-2a^2b+ab^2-a^2b+2ab^2-b^3\\
& \colMM{red}{\Darr 同類項を整理}\\
&= a^3-3a^2b+3ab^2-b^3 \colMM{red}{\cdots 完成！}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan\def\Left{a}
\def\Right{(-b)}
\def\A{a^3}
\def\B{3a^2b}
\def\C{3ab^2}
\def\D{b^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& (a-b)^3\\
& \colMM{red}{ 左\ \ +\ 右 に書き換える}\\
&= \left\{\colBX{bisque}{$\Left$}+\colBX{palegreen}{$\Right$}\right\}^3\\
& \colMM{red}{左^3 +3\,左^2\,右^1 +3\,左^1\,右^2 +右^3}\\
&= (\colBX{bisque}{$\Left$})^3+3(\colBX{bisque}{$\Left$})^2\colBX{palegreen}{$\Right$}^1+3(\colBX{bisque}{$\Left$})^1\colBX{palegreen}{$\Right$}^2+\colBX{palegreen}{$\Right$}^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan\def\Left{a}
\def\Right{b}
\def\A{a^3}
\def\B{3a^2b}
\def\C{3ab^2}
\def\D{b^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan次の式を展開しなさい。
↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{1}
\def\A{x^3}
\def\B{3x^2}
\def\C{3x}
\def\D{1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{2x}
\def\Right{y}
\def\A{8x^3}
\def\B{12x^2y}
\def\C{6xy^2}
\def\D{y^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{2}
\def\A{x^3}
\def\B{6x^2}
\def\C{12x}
\def\D{8}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{3a}
\def\Right{b}
\def\A{27a^3}
\def\B{27a^2b}
\def\C{9ab^2}
\def\D{b^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan↓この問題へのリンクはこちら（右クリックで保存）
【解答】
\def\Left{x}
\def\Right{2y}
\def\A{x^3}
\def\B{6x^2y}
\def\C{12xy^2}
\def\D{8y^3}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
& \colMM{red}{ 左\ \ -\ 右}\\
& (\colBX{bisque}{$\Left$}-\colBX{palegreen}{$\Right$})^3\\
& \colMM{red}{左^3 -3\,左^2\,右^1 +3\,左^1\,右^2 -右^3}\\
&= (\colBX{bisque}{$\Left$})^3-3(\colBX{bisque}{$\Left$})^2(\colBX{palegreen}{$\Right$})^1+3(\colBX{bisque}{$\Left$})^1(\colBX{palegreen}{$\Right$})^2-(\colBX{palegreen}{$\Right$})^3\\
\\
&= \A - \B + \C - \D
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan
