$(3a-2b-c)^2$ を展開せよ。代数学展開多項式因数分解2025/5/171. 問題の内容(3a−2b−c)2(3a-2b-c)^2(3a−2b−c)2 を展開せよ。2. 解き方の手順(3a−2b−c)2(3a-2b-c)^2(3a−2b−c)2 は (3a−2b−c)(3a−2b−c)(3a-2b-c)(3a-2b-c)(3a−2b−c)(3a−2b−c) と同じです。これを展開します。まず、3a3a3a を分配します。3a(3a−2b−c)=9a2−6ab−3ac3a(3a-2b-c) = 9a^2 - 6ab - 3ac3a(3a−2b−c)=9a2−6ab−3ac次に、−2b-2b−2b を分配します。−2b(3a−2b−c)=−6ab+4b2+2bc-2b(3a-2b-c) = -6ab + 4b^2 + 2bc−2b(3a−2b−c)=−6ab+4b2+2bc最後に、−c-c−c を分配します。−c(3a−2b−c)=−3ac+2bc+c2-c(3a-2b-c) = -3ac + 2bc + c^2−c(3a−2b−c)=−3ac+2bc+c2これらをすべて足し合わせます。9a2−6ab−3ac−6ab+4b2+2bc−3ac+2bc+c29a^2 - 6ab - 3ac - 6ab + 4b^2 + 2bc - 3ac + 2bc + c^29a2−6ab−3ac−6ab+4b2+2bc−3ac+2bc+c2同類項をまとめます。9a2+4b2+c2−6ab−6ab−3ac−3ac+2bc+2bc9a^2 + 4b^2 + c^2 - 6ab - 6ab - 3ac - 3ac + 2bc + 2bc9a2+4b2+c2−6ab−6ab−3ac−3ac+2bc+2bc9a2+4b2+c2−12ab−6ac+4bc9a^2 + 4b^2 + c^2 - 12ab - 6ac + 4bc9a2+4b2+c2−12ab−6ac+4bc3. 最終的な答え9a2+4b2+c2−12ab−6ac+4bc9a^2 + 4b^2 + c^2 - 12ab - 6ac + 4bc9a2+4b2+c2−12ab−6ac+4bc