【数学I】例題1.1.15：因数分解（a^3+b^3+c^3-3abcの形）（One More）★★★

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% 例題I1.1.15：因数分解（$a^3+b^3+c^3-3abc$の形）（One More）★★★
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(1) $a^3+b^3=(a+b)^3-3ab(a+b)$であることを用いて，$a^3+b^3+c^3-3abc$を因数分解せよ．

(2) $x^3+y^3+3xy-1$を因数分解せよ．

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% 例題I1.1.15の解答（One More）★★★
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(1)

\begin{align*}
&a^3+b^3+c^3-3abc\\
=&(a+b)^3-3ab(a+b)+c^3-3abc\\
=&\{(a+b)^3+c^3\}-3ab(a+b)-3abc\\
=&(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)\\
=&(a+b+c)\{(a+b)^2-(a+b)c+c^2-3ab\}\\
=&(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)\\
=&(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\end{align*}

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% 例題I1.1.15の別解
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\begin{align*}
&a^3+b^3+c^3-3abc\\
=&(a+b)^3-3ab(a+b)+c^3-3abc\\
=&\{(a+b)^3+c^3\}-3ab(a+b)-3abc\\
=&\{(a+b)+c\}^3-3(a+b)c\{(a+b)+c\}-3ab(a+b+c)\\
=&(a+b+c)\{(a+b+c)^2-3(a+b)c-3ab\}\\
=&(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\end{align*}

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% 例題I1.1.15の解答（One More）★★★
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(2)

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

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% 問題I1.1.15：因数分解（$a^3+b^3+c^3-3abc$の形）（One More）★★★
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次の式を因数分解せよ．

(1) $p^3+q^3+3pq-1$

(2) $(x-y)^3+(y-z)^3+(z-x)^3$

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% 問題I1.1.15の解答（One More）★★★
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(1)

\begin{align*}
&p^3+q^3+(-1)^3+3pq\\
=&p^3+q^3+(-1)^3-3p\cdot q\cdot(-1)\\
=&\{p+q+(-1)\}\{p^2+q^2+(-1)^2-p\cdot q-q\cdot(-1)-(-1)\cdot p\}\\
=&(p+q-1)(p^2+q^2-pq+p+q+1)
\end{align*}

(2) $x-y=a,y-z=b,z-x=c$とおく．

$a+b+c=(x-y)+(y-z)+(z-x)=0$より，

\begin{align*}
&(x-y)^3+(y-z)^3+(z-x)^3\\
&=a^3+b^3+c^3\\
&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc\\
&=3abc\\
&=3(x-y)(y-z)(z-x)
\end{align*}

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% 例題I1.1.15の側注：因数分解（$a^3+b^3+c^3-3abc$の形）（One More）★★★
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% 側注1
\begin{align*}
&a^3+b^3\\
&=(a+b)^3-3ab(a+b)
\end{align*}

% 側注2
$(a+b)^3+c^3$に3次式の因数分解の公式を適用する．

% 側注3
$a+b+c$が共通因数となる．

% 側注4
輪環の順に整理する．

% 側注5
$a+b=A$とおくと，

\begin{align*}
&A^3+c^3\\
=&(A+c)^3-3Ac(A+c)
\end{align*}

% 側注6
(1)において，$a=x,b=y,c=-1$の場合である（それぞれ代入する）．

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% 問題I1.1.15の側注：因数分解（$a^3+b^3+c^3-3abc$の形）（One More）★★★
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% 側注1
$a+b+c=0$

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% 例題I1.1.15の考え方・補足など：因数分解（$a^3+b^3+c^3-3abc$の形）（One More）★★★
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% 考え方
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$a^3+b^3=(a+b)^3-3ab(a+b)\cdots(\mathrm{i})$を用いて変形すると，

\[
a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc=(a+b)^3+c^3-3ab\{(a+b)+c\}
\]

ここで，$(a+b)^3+c^3$について3次式の因数分解の公式$a^3+b^3=(a+b)(a^2-ab+b^2)$か(i)を適用すると，さらに因数分解することができる．

(2)は，(1)で得られた式$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$を用いる．

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% 因数分解の公式（3次式）
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\[
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\]