【数学I】例題1.1.14：因数分解（対称式，交代式）（One More）★★

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% 例題I1.1.14：因数分解（対称式，交代式）（One More）★★
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次の式を因数分解せよ．

(1) $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+2abc$

(2) $a^2(b-c)+b^2(c-a)+c^2(a-b)$

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% 例題I1.1.14の解答（One More）★★
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(1)

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

(2)

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

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% 問題I1.1.14：因数分解（対称式，交代式）（One More）★★
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次の式を因数分解せよ．

(1) $(a+b)(b+c)(c+a)+abc$

(2) $ab(a-b)+bc(b-c)+ca(c-a)$

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% 問題I1.1.14の解答（One More）★★
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(1)

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

(2)

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

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% 例題I1.1.14の側注：因数分解（対称式，交代式）（One More）★★
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% 側注1
$a$について整理する．

% 側注2
$b+c$をくくり出す．

% 側注3
このままでも正答であるが，輪環の順に整理するとよい．

% 側注4
$a$について整理する．

% 側注5
$b-c$をくくり出す．

% 側注6
このままでも正答であるが，輪環の順に整理するとよい．

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% 問題I1.1.14の側注：因数分解（対称式，交代式）（One More）★★
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% 側注1
$a$について整理する．

% 側注2
たすき掛けを用いて因数分解する．

% 側注3
$a$について整理する．

% 側注4
$b-c$をくくり出す．

% 側注5
このままでも正答であるが，輪環の順に整理するとよい．

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% 例題I1.1.14の考え方・補足など：因数分解（対称式，交代式）（One More）★★
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% 考え方
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$a,b,c$いずれについても2次式であるから，例えば$a$について整理する．また，答えを記すときは，慣習として輪環の順に整理することが多い．

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% 余談
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$a,b,c$の多項式で，$a,b,c$のどの2つの文字を入れ替えても，もとの式と同じ式になるものを対称式という．

また，$a,b,c$の多項式で，$a,b,c$のどの2つの文字を入れ替えても，もとの式と符号だけ変わる式になるものを交代式という．

例えば，2つの文字$a,b$の対称式は$a^2+b^2,a^3+b^3$，交代式は$a-b,a^2-b^2$などがある．

さらに，(2)のような3つの文字の交代式は，最簡（さいかん）交代式といわれる$(a-b)(b-c)(c-a)$を因数にもつことが知られている．