
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