【数学I】例題1.1.6：おき換えを用いた展開（One More）★★

% 例題I1.1.6：おき換えを用いた展開 （One More）★★
次の式を展開せよ. (1)$(x-y+2)(x-y-4)$(2)$(3x+4y-z)(3x-4y+z)$(3)$(a+b-c+d)(a+b+c-d)$

% 解答（例題I1.1.6）
(1)$\begin{aligned} &(x-y+2)(x-y-4)\\ =& \{(x-y)+2\}\{(x-y)-4\} \\ =&(x-y)^2-2(x-y)-8 \\ =& x^2-2xy+y^2-2x+2y-8 \end{aligned}$(2)$\begin{aligned} &(3x+4y-z)(3x-4y+z)\\ =& \{3x+(4y-z)\}\{3x-(4y-z)\} \\ =&(3x)^2-(4y-z)^2 \\ =& 9x^2-\left(16y^2-8yz+z^2\right)\\ =& 9x^2-16y^2+8yz-z^2 \end{aligned}$(3)$\begin{aligned} &(a+b+c-d)(a+b-c+d)\\ =& \{(a+b)+(c-d)\}\{(a+b)-(c-d)\} \\ =&(a+b)^2-(c-d)^2 \\ =& \left(a^2+2ab+b^2\right)-\left(c^2-2cd+d^2\right)\\ =& a^2+b^2-c^2-d^2+2ab+2cd \end{aligned}$

% 問題I1.1.6
次の式を展開せよ. (1)$(x+2y-3)(x+2y+5)$(2)$(4a-3b+c)(4a+3b-c)$(3)$(p+q-r+s)(p+q+r-s)$

% 解答I1.1.6
(1)$\begin{aligned} &(x+2y-3)(x+2y+5)\\ =& \{(x+2y)-3\}\{(x+2y)+5\} \\ =&(x+2y)^2+2(x+2y)-15 \\ =& x^2+4xy+4y^2+2x+4y-15 \end{aligned}$(2)$\begin{aligned} &(4a-3b+c)(4a+3b-c)\\ =& \{4a+(-3b+c)\}\{4a-(-3b+c)\} \\ =&(4a)^2-(-3b+c)^2 \\ =& 16a^2-\left(9b^2-6bc+c^2\right)\\ =& 16a^2-9b^2+6bc-c^2 \end{aligned}$(3)$\begin{aligned} &(p+q-r+s)(p+q+r-s)\\ =& \{(p+q)+(r-s)\}\{(p+q)-(r-s)\} \\ =&(p+q)^2-(r-s)^2 \\ =& \left(p^2+2pq+q^2\right)-\left(r^2-2rs+s^2\right)\\ =& p^2+q^2-r^2-s^2+2pq+2rs \end{aligned}$