【数学I】例題1.1.7：掛ける順序や組み合わせを工夫した展開（One More）★★

% 例題I1.1.7：掛ける順序や組み合わせを工夫した展開 （One More）★★
次の式を展開せよ. (1)$(x-1)(x-4)(x+3)(x+6)$(2)$(x-2)(x+2)\left(x^2+4\right)\left(x^4+4\right)$(3)$(p-q)^2(p+q)^2\left(p^2+q^2\right)^2$

% 解答（例題I1.1.7）
(1)$\begin{aligned} &(x-1)(x-4)(x+3)(x+6)\\ =&(x-1)(x+3)\times(x-4)(x+6)\\ =& \left(x^2+2x-3\right)\left(x^2+2x-24\right)\\ =& \left\{\left(x^2+2x\right)-3\right\}\left\{\left(x^2+2x\right)-24\right\} \\ =& \left(x^2+2x\right)^2-27\left(x^2+2x\right)+72 \\ =& x^4+4x^3+4x^2-27x^2-54x+72 \\ =& x^4+4x^3-23x^2-54x+72 \end{aligned}$(2)$\begin{aligned} &(x-2)(x+2)\left(x^2+4\right)\left(x^4+4\right)\\ =& \left(x^2-4\right)\left(x^2+4\right)\left(x^4+4\right)\\ =& \left(x^4-16\right)\left(x^4+4\right)\\ =& x^8+4x^4-16x^4-64 \\ =& x^8-12x^4-64 \end{aligned}$(3)$\begin{aligned} &(p-q)^2(p+q)^2\left(p^2+q^2\right)^2 \\ =& \left\{(p-q)(p+q)\left(p^2+q^2\right)\right\}^2\\ =& \left\{\left(p^2-q^2\right)\left(p^2+q^2\right)\right\}^2 \\ =& \left\{\left(p^2\right)^2-\left(q^2\right)^2\right\}^2 \\ =& \left(p^4-q^4\right)^2\\ =&p^8-2p^4q^4+q^8 \end{aligned}$

% 問題I1.1.7
次の式を展開せよ. (1)$(x-4)(x-5)(x+2)(x+3)$(2)$(x-3)(x+3)\left(x^2+9\right)\left(x^4+9\right)$(3)$(m-n)^2\left(m^2+mn+n^2\right)^2$

% 解答I1.1.7
(1)$\begin{aligned} &(x-4)(x-5)(x+2)(x+3)\\ =&(x-4)(x+2)\times(x-5)(x+3)\\ =& \left(x^2-2x-8\right)\left(x^2-2x-15\right)\\ =& \left\{\left(x^2-2x\right)-8\right\}\left\{\left(x^2-2x\right)-15\right\} \\ =& \left(x^2-2x\right)^2-23\left(x^2-2x\right)+120 \\ =& x^4-4x^3+4x^2-23x^2+46x+120 \\ =& x^4-4x^3-19x^2+46x+120 \end{aligned}$(2)$\begin{aligned} &(x-3)(x+3)\left(x^2+9\right)\left(x^4+9\right)\\ =& \left(x^2-9\right)\left(x^2+9\right)\left(x^4+9\right)\\ =& \left(x^4-81\right)\left(x^4+9\right)\\ =& x^8+9x^4-81x^4-729 \\ =& x^8-72x^4-729 \end{aligned}$(3)$\begin{aligned} &(m-n)^2\left(m^2+mn+n^2\right)^2 \\ =& \left\{(m-n)\left(m^2+mn+n^2\right)\right\}^2\\ =&\left(m^3-n^3\right)^2\\ =&m^6-2m^3n^3+n^6 \end{aligned}$