【数学I】例題1.1.5：乗法公式（3次）を用いた展開（One More）★★

% 例題I1.1.5：乗法公式（3次）を用いた展開 （One More）★★
次の式を展開せよ. (1)$(2x+1)^3$(2)$(3x-4y)^3$(3)$(2x+1)(4x^2-2x+1)$(4)$(4a-3b)(16a^2+12ab+9b^2)$(5)$(x+2)^3(x-2)^3$(6)$(x+1)(x-1)(x^2+x+1)(x^2-x+1)$

% 解答（例題I1.1.5）
(1)$\begin{aligned} (2x+1)^3 &=(2x)^3+3(2x)^2 \cdot 1+3(2x) \cdot 1^2+1^3\\ &=8x^3+12x^2+6x+1 \end{aligned}$(2)$\begin{aligned} (3x-4y)^3 &=(3x)^3-3(3x)^2 \cdot 4y+3 \cdot 3x \cdot (4y)^2-(4y)^3 \\ &=27x^3-108x^2y+144xy^2-64y^3 \end{aligned}$(3)$\begin{aligned} (2x+1)(4x^2-2x+1)&=(2x+1)\{(2x)^2-2x \cdot 1+1^2\} \\ &=(2x)^3+1^3 \\ &=8x^3+1 \end{aligned}$(4)$\begin{aligned} (4a-3b)(16a^2+12ab+9b^2)&=(4a-3b)\{(4a)^2+4a \cdot 3b+(3b)^2\} \\ &=(4a)^3-(3b)^3 \\ &=64a^3-27b^3 \end{aligned}$(5)$\begin{aligned} (x+2)^3(x-2)^3 &=\{(x+2)(x-2)\}^3 \\ &=(x^2-4)^3 \\ &=(x^2)^3-3(x^2)^2 \cdot 4+3x^2 \cdot 4^2-4^3 \\ &=x^6-12x^4+48x^2-64 \end{aligned}$(6)$\begin{aligned} &(x+1)(x-1)(x^2+x+1)(x^2-x+1)\\ =&(x+1)(x^2-x+1)(x-1)(x^2+x+1)\\ =&(x^3+1)(x^3-1)\\ =& x^6-1 \end{aligned}$

% 問題I1.1.5
次の式を展開せよ. (1)$(x+3)^3$(2)$(2x-5y)^3$(3)$(3x-2)(9x^2+6x+4)$(4)$(5a+2b)(25a^2-10ab+4b^2)$(5)$(x-3)^3(x+3)^3$(6)$(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)$

% 解答I1.1.5
(1)$\begin{aligned} (x+3)^3 &=x^3+3x^2 \cdot 3+3x \cdot 3^2+3^3 \\ &=x^3+9x^2+27x+27 \end{aligned}$(2)$\begin{aligned} (2x-5y)^3 &=(2x)^3-3(2x)^2 \cdot 5y+3 \cdot 2x \cdot (5y)^2-(5y)^3 \\ &=8x^3-60x^2y+150xy^2-125y^3 \end{aligned}$(3)$\begin{aligned} (3x-2)(9x^2+6x+4)&=(3x-2)\{(3x)^2+3x \cdot 2+2^2\} \\ &=(3x)^3-2^3 \\ &=27x^3-8 \end{aligned}$(4)$\begin{aligned} (5a+2b)(25a^2-10ab+4b^2)&=(5a+2b)\{(5a)^2-5a \cdot 2b+(2b)^2\} \\ &=(5a)^3+(2b)^3 \\ &=125a^3+8b^3 \end{aligned}$(5)$\begin{aligned} (x-3)^3(x+3)^3 &=\{(x-3)(x+3)\}^3 \\ &=(x^2-9)^3 \\ &=(x^2)^3-3(x^2)^2 \cdot 9+3x^2 \cdot 9^2-9^3 \\ &=x^6-27x^4+243x^2-729 \end{aligned}$(6)$\begin{aligned} &(x+2)(x-2)(x^2+2x+4)(x^2-2x+4)\\ =&(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)\\ =&(x^3+8)(x^3-8)\\ =& x^6-64 \end{aligned}$