【数学I】例題1.1.5：乗法公式（3次）を用いた展開（One More）★★

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% 例題I1.1.5：乗法公式（3次）を用いた展開（One More）★★
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次の式を展開せよ．

(1) $(2x+1)^3$

(2) $(3x-4y)^3$

(3) $(2x+1)(4x^2-2x+1)$

(4) $(4a-3b)(16a^2+12ab+9b^2)$

(5) $(x+2)^3(x-2)^3$

(6) $(x+1)(x-1)(x^2+x+1)(x^2-x+1)$

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% 例題I1.1.5の解答（One More）★★
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(1)

\begin{align*}
(2x+1)^3&=(2x)^3+3(2x)^2\cdot 1+3(2x)\cdot 1^2+1^3\\
&=8x^3+12x^2+6x+1
\end{align*}

(2)

\begin{align*}
(3x-4y)^3&=(3x)^3-3(3x)^2\cdot 4y+3\cdot 3x\cdot(4y)^2-(4y)^3\\
&=27x^3-108x^2y+144xy^2-64y^3
\end{align*}

(3)

\begin{align*}
(2x+1)(4x^2-2x+1)&=(2x+1)\{(2x)^2-2x\cdot 1+1^2\}\\
&=(2x)^3+1^3\\
&=8x^3+1
\end{align*}

(4)

\begin{align*}
(4a-3b)(16a^2+12ab+9b^2)&=(4a-3b)\{(4a)^2+4a\cdot 3b+(3b)^2\}\\
&=(4a)^3-(3b)^3\\
&=64a^3-27b^3
\end{align*}

(5)

\begin{align*}
(x+2)^3(x-2)^3&=\{(x+2)(x-2)\}^3\\
&=(x^2-4)^3\\
&=(x^2)^3-3(x^2)^2\cdot 4+3x^2\cdot 4^2-4^3\\
&=x^6-12x^4+48x^2-64
\end{align*}

(6)

\begin{align*}
&(x+1)(x-1)(x^2+x+1)(x^2-x+1)\\
=&(x+1)(x^2-x+1)(x-1)(x^2+x+1)\\
=&(x^3+1)(x^3-1)\\
=&x^6-1
\end{align*}

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% 問題I1.1.5：乗法公式（3次）を用いた展開（One More）★★
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次の式を展開せよ．

(1) $(x+3)^3$

(2) $(2x-5y)^3$

(3) $(3x-2)(9x^2+6x+4)$

(4) $(5a+2b)(25a^2-10ab+4b^2)$

(5) $(x-3)^3(x+3)^3$

(6) $(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)$

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% 問題I1.1.5の解答（One More）★★
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(1)

\begin{align*}
(x+3)^3&=x^3+3x^2\cdot 3+3x\cdot 3^2+3^3\\
&=x^3+9x^2+27x+27
\end{align*}

(2)

\begin{align*}
(2x-5y)^3&=(2x)^3-3(2x)^2\cdot 5y+3\cdot 2x\cdot(5y)^2-(5y)^3\\
&=8x^3-60x^2y+150xy^2-125y^3
\end{align*}

(3)

\begin{align*}
(3x-2)(9x^2+6x+4)&=(3x-2)\{(3x)^2+3x\cdot 2+2^2\}\\
&=(3x)^3-2^3\\
&=27x^3-8
\end{align*}

(4)

\begin{align*}
(5a+2b)(25a^2-10ab+4b^2)&=(5a+2b)\{(5a)^2-5a\cdot 2b+(2b)^2\}\\
&=(5a)^3+(2b)^3\\
&=125a^3+8b^3
\end{align*}

(5)

\begin{align*}
(x-3)^3(x+3)^3&=\{(x-3)(x+3)\}^3\\
&=(x^2-9)^3\\
&=(x^2)^3-3(x^2)^2\cdot 9+3x^2\cdot 9^2-9^3\\
&=x^6-27x^4+243x^2-729
\end{align*}

(6)

\begin{align*}
&(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)\\
=&(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)\\
=&(x^3+8)(x^3-8)\\
=&x^6-64
\end{align*}

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% 例題I1.1.5の側注：乗法公式（3次）を用いた展開（One More）★★
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% 側注1
\begin{align*}
&(a+b)^3\\
=&a^3+3a^2b+3ab^2+b^3
\end{align*}

% 側注2
\begin{align*}
&(a-b)^3\\
=&a^3-3a^2b+3ab^2-b^3
\end{align*}

% 側注3
\begin{align*}
&(a+b)(a^2-ab+b^2)\\
=&a^3+b^3
\end{align*}

% 側注4
\begin{align*}
&(a-b)(a^2+ab+b^2)\\
=&a^3-b^3
\end{align*}

% 側注5
先に3次の乗法公式を用いて展開してもよいが，計算に手間が掛かる．そこで，$(a+b)(a-b)=a^2-b^2$の利用を考える．

% 側注6
$(a+b)(a-b)=a^2-b^2$

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% 問題I1.1.5の側注：乗法公式（3次）を用いた展開（One More）★★
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% 側注1
\begin{align*}
&(a+b)^3\\
=&a^3+3a^2b+3ab^2+b^3
\end{align*}

% 側注2
\begin{align*}
&(a-b)^3\\
=&a^3-3a^2b+3ab^2-b^3
\end{align*}

% 側注3
\begin{align*}
&(a+b)(a^2-ab+b^2)\\
=&a^3+b^3
\end{align*}

% 側注4
\begin{align*}
&(a-b)(a^2+ab+b^2)\\
=&a^3-b^3
\end{align*}

% 側注5
先に3次の乗法公式を用いて展開してもよいが，計算に手間が掛かる．そこで，$(a+b)(a-b)=a^2-b^2$の利用を考える．

% 側注6
$(a+b)(a-b)=a^2-b^2$

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% 例題I1.1.5の考え方・補足など：乗法公式（3次）を用いた展開（One More）★★
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% 考え方
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3次の乗法公式を用いるときは，符号に注意すること．

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% 3次の乗法公式
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\[
(a+b)^3=a^3+3a^2b+3ab^2+b^3,(a-b)^3=a^3-3a^2b+3ab^2-b^3,
\]

\[
(a+b)(a^2-ab+b^2)=a^3+b^3,(a-b)(a^2+ab+b^2)=a^3-b^3
\]