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\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+2abc+b^2c+bc^2\right)\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+abc+abc+b^2c+bc^2\right)\)
\(=3\left[ab\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=3\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(M=a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(M=ab^2-ac^2+bc^2-ba^2+c\left(a-b\right)\left(a+b\right)\)
\(M=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a-b\right)\left(a+b\right)\)
\(M=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(M=\left(a-b\right)\left[-a\left(b-c\right)+c\left(b-c\right)\right]\)
\(M=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Giờ là cách khác:(tại em làm khá kĩ nên nó dài thôi chứ em trình bày lại trong giấy nó ngắn ngủn à)
Đặt \(b^2-c^2=x;c^2-a^2=y\Rightarrow a^2-b^2=-\left(x+y\right)\)
Suy ra \(M=ax+by-c\left(x+y\right)\)
\(=x\left(a-c\right)+y\left(b-c\right)\)
\(=\left(b^2-c^2\right)\left(a-c\right)+\left(c^2-a^2\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(b+c\right)+\left(c-a\right)\left(b-c\right)\left(c+a\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(b+c\right)-\left(a-c\right)\left(b-c\right)\left(c+a\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(b+c-c-a\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(b-a\right)\) [muốn cho đẹp thì nhân (-1) . (-1) vào thì nó thành (a-b)(b-c)(c-a) ]
\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-4c^2\)
\(=2\left(a+b\right)^2-2c^2=2\left[\left(a+b\right)^2-c^2\right]=2\left(a+b+c\right)\left(a+b-c\right)\)
Mình đã làm bài này rồi.
Link: https://hoc24.vn/hoi-dap/question/824554.html
\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2\)
\(=\left[\left(a+b+c\right)^2-\left(2c\right)^2\right]+\left(a+b-c\right)^2\)
\(=\left(a+b+3c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\)
\(=\left(a+b-c\right)\left(a+b+3c+a+b-c\right)\)
\(=\left(a+b-c\right)\left(2a+2b+2c\right)\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
=(a-b)(b-c)(c-a)
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left[\left(b-a\right)-\left(c-a\right)\right]+ca\left(c-a\right)\)
\(=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(=\left(a-b\right)\left(ab-bc\right)-\left(c-a\right)\left(bc-ca\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(c-a\right)\left(b-a\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(a-c\right)\left(a-b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)