以下の2つの式を因数分解する問題です。 (1) $a^3b + 16 - 4ab - 4a^2$ (2) $x^3y + x^2 - xyz^2 - z^2$代数学因数分解多項式式変形2025/4/291. 問題の内容以下の2つの式を因数分解する問題です。(1) a3b+16−4ab−4a2a^3b + 16 - 4ab - 4a^2a3b+16−4ab−4a2(2) x3y+x2−xyz2−z2x^3y + x^2 - xyz^2 - z^2x3y+x2−xyz2−z22. 解き方の手順(1) a3b+16−4ab−4a2a^3b + 16 - 4ab - 4a^2a3b+16−4ab−4a2まずは項の順番を整理します。a3b−4a2−4ab+16a^3b - 4a^2 - 4ab + 16a3b−4a2−4ab+16a2(ab−4)−4(ab−4)a^2(ab - 4) - 4(ab - 4)a2(ab−4)−4(ab−4)(a2−4)(ab−4)(a^2 - 4)(ab - 4)(a2−4)(ab−4)(a−2)(a+2)(ab−4)(a - 2)(a + 2)(ab - 4)(a−2)(a+2)(ab−4)(2) x3y+x2−xyz2−z2x^3y + x^2 - xyz^2 - z^2x3y+x2−xyz2−z2x3y−xyz2+x2−z2x^3y - xyz^2 + x^2 - z^2x3y−xyz2+x2−z2xy(x2−z2)+(x2−z2)xy(x^2 - z^2) + (x^2 - z^2)xy(x2−z2)+(x2−z2)(xy+1)(x2−z2)(xy + 1)(x^2 - z^2)(xy+1)(x2−z2)(xy+1)(x−z)(x+z)(xy + 1)(x - z)(x + z)(xy+1)(x−z)(x+z)3. 最終的な答え(1) (a−2)(a+2)(ab−4)(a - 2)(a + 2)(ab - 4)(a−2)(a+2)(ab−4)(2) (xy+1)(x−z)(x+z)(xy + 1)(x - z)(x + z)(xy+1)(x−z)(x+z)