%2007/04/22 06:10
%Copyright 2007 WATANABE,Masayuki
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\begin{document}
{\large 京都大学　1974年　入学試験　文系数学　問題3}\\
\\
{\large 問題}\\
\\
$(x^3+\sqrt{2}\,x^2+\sqrt[3]{3}\,x+1)^{100}$ を展開したときの，$x^{296}$ の係数を求めよ．\\
\\
{\large 解答}\\
\\
$x^3+\sqrt{2}\,x^2+\sqrt[3]{3}\,x+1=0$の解を$(a,b,c)$とすると、\\
$x$の３次の係数が１なので\\
$x^3+\sqrt{2}\,x^2+\sqrt[3]{3}\,x+1=(x-a)(x-b)(x-c)$と分解される。
このとき、
$(x^3+\sqrt{2}\,x^2+\sqrt[3]{3}\,x+1)^{100}$をかんがえると\\
$(x^3+\sqrt{2}\,x^2+\sqrt[3]{3}\,x+1)^{100}=((x-a)(x-b)(x-c))^{100}$\\
これを展開すると、$x^{296}$の係数は
$-a,-b,-c$を$300-296=4$回使って作れる組み合わせの和\\
つまり、\\
$a,a,a,a,{}_{100}\text{C}_4=\dfrac{100 \cdot 99 \cdot 98 \cdot 97}{4 \cdot 3 \cdot 2}=3921225$通り\\
$a,a,a,b,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\
$a,a,a,c,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\
$a,a,b,b,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_2=\dfrac{100 \cdot 99}{2}\cdot \dfrac{100 \cdot 99}{2}=24502500$通り\\
$a,a,b,c,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_1 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99}{2}\cdot 100\cdot 100=49500000$通り\\


$a,a,c,c,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_2=\dfrac{100 \cdot 99}{2}\cdot \dfrac{100 \cdot 99}{2}=24502500$通り\\
$a,b,b,b,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\
$a,b,b,c,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_1 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99}{2}\cdot 100\cdot 100=49500000$通り\\
$a,b,c,c,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_1 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99}{2}\cdot 100\cdot 100=49500000$通り\\
$a,c,c,c,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\
$b,b,b,b,{}_{100}\text{C}_4=\dfrac{100 \cdot 99 \cdot 98 \cdot 97}{4 \cdot 3 \cdot 2}=3921225$通り\\
$b,b,b,c,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\

$b,b,c,c,{}_{100}\text{C}_2 \cdot {}_{100}\text{C}_2=\dfrac{100 \cdot 99}{2}\cdot \dfrac{100 \cdot 99}{2}=24502500$通り\\
$b,c,c,c,{}_{100}\text{C}_3 \cdot {}_{100}\text{C}_1=\dfrac{100 \cdot 99 \cdot 98}{3 \cdot 2}\cdot 100=16170000$通り\\
$c,c,c,c,{}_{100}\text{C}_4=\dfrac{100 \cdot 99 \cdot 98 \cdot 97}{4 \cdot 3 \cdot 2}=3921225$通り\\
よって係数は\\

$ 3921225 a^4
+16170000 a^3b
+16170000 a^3c
+24502500 a^2b^2
+49500000 a^2bc\\
+24502500 a^2c^2
+16170000 ab^3
+49500000 ab^2c
+49500000 abc^2
+16170000 ac^3\\
+ 3921225 b^4
+16170000 b^3c
+24502500 b^2c^2
+16170000 bc^3
+ 3921225 c^4$\\
\\
\begin{eqnarray*}
\lefteqn{
 3921225 a^4
+16170000 a^3b
+16170000 a^3c
+24502500 a^2b^2
+49500000 a^2bc}\\
& &+24502500 a^2c^2
+16170000 ab^3
+49500000 ab^2c
+49500000 abc^2
+16170000 ac^3\\
& &+ 3921225 b^4
+16170000 b^3c
+24502500 b^2c^2
+16170000 bc^3
+ 3921225 c^4\\
&=&
  3921225(a^4+b^4+c^4)+\\
& &+16170000(a^3b+a^3c+ab^3+ac^3+b^3c+bc^3)\\
& &+24502500(a^2b^2+a^2c^2+b^2c^2)+\\
& &+49500000(ab^2c+abc^2+a^2bc)\\
&=&
  3921225(a^4+b^4+c^4)\\
& &+16170000(a^3b+a^3c+ab^3+ac^3+b^3c+bc^3)\\
& &+24502500(a^2b^2+a^2c^2+b^2c^2)\\
& &+49500000(abc)(a+b+c)\\
\end{eqnarray*}
\begin{eqnarray*}
abc&=&-1\\
a+b+c&=&-\sqrt{2}\\
ab+bc+ac&=&\sqrt[3]{3}\\
a^2+b^2+c^2&=&(a+b+c)^2-2(ab+ac+bc)\\
&=&2-2\sqrt[3]{3}\\
\end{eqnarray*}
\begin{eqnarray*}
\lefteqn{a^2b^2+a^2c^2+b^2c^2}\\
&=&(ab+bc+ca)^2-2(abc)(a+b+c)\\
&=&\sqrt[3]{3}^2-2\sqrt{2}\\
\lefteqn{a^4+b^4+c^4}\\
&=&-8(abc)(a+b+c)+(a+b+c)^4-4(ab+bc+ca)(a^2+b^2+c^2)-6(a^2b^2+a^2c^2+b^2c^2)\\
&=&-8(-1)(-\sqrt{2})+(-\sqrt{2})^4-4(\sqrt[3]{3})(2-2\sqrt[3]{3})-6(\sqrt[3]{3}^2-2\sqrt{2})\\
&=&2\sqrt[3]{3}^2-8\sqrt[3]{3}+4\sqrt{2}+4\\
\lefteqn{a^3b+a^3c+ab^3+ac^3+b^3c+bc^3}\\
&=&(ab+bc+ca)(a^2+b^2+c^2)-abc(a+b+c)\\
&=&(\sqrt[3]{3})(2-2\sqrt[3]{3})-(\sqrt{2})\\
&=&-2\sqrt[3]{3}^2+2\sqrt[3]{3}-\sqrt{2}
\end{eqnarray*}
より
\begin{eqnarray*}
& &3921225(a^4+b^4+c^4)\\
& &+16170000(a^3b+a^3c+ab^3+ac^3+b^3c+bc^3)\\
& &+24502500(a^2b^2+a^2c^2+b^2c^2)\\
& &+49500000(abc)(a+b+c)\\
&=&
  3921225(2\sqrt[3]{3}^2-8\sqrt[3]{3}+4\sqrt{2}+4)\\
& &+16170000(-2\sqrt[3]{3}^2+2\sqrt[3]{3}-\sqrt{2})\\
& &+24502500(\sqrt[3]{3}^2-2\sqrt{2})\\
& &+49500000(\sqrt{2})\\
&=&4950 \sqrt[3]{3}^2 +970200\sqrt[3]{3}+9900\sqrt{2}+ 15684900
\end{eqnarray*}
$$4950 \sqrt[3]{3}^2 +970200\sqrt[3]{3}+9900\sqrt{2}+ 15684900$$
\end{document}




