\(a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2\left(c-a\right)\)

\(=a^2b^2\left(a-b\right)-b^2c^2\left(a-b\right)+c^2a^2\left(c-a\right)-b^2c^2\left(c-a\right)\)

\(=\left(a-b\right)b^2\left(a-c\right)\left(a+c\right)+\left(c-a\right)c^2\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(a-c\right)\left(ab^2+cb^2-c^2a-c^2b\right)\)

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)

t làm bên h rồi mà? Làm quá lâu rồi luôn ấy! Đáp án y chang bạn Kid:v

Câu hỏi của Trần Minh Hiển - Toán lớp 9 (không biết AD đã fix lỗi ko dán link h vào olm chưa, nếu chưa ib t gửi full link, nhớ kèm theo link câu hỏi này là ok.)

\(3\left(a+3b\right)\left(b+3c\right)\left(c+3a\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)

\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)

\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)

\(\Rightarrow B=x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)

\(x+y=a+b-2c+b+c-2a=2b-a-c\)

\(y+z=b+c-2a+c+a-2b=2c-a-b\)

\(z+x=c+a-2b+a+b-2c=2a-b-c\)

Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)

Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)

Lời giải :

\(B=2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)

\(B=2b^2c+4bc^2+2ac^2-4a^2c-2ab\left(a+2b\right)-7abc\)

\(B=abc+2b^2c-4a^2c-8abc-2ab\left(a+2b\right)+2ac^2+4bc^2\)

\(B=bc\left(a+2b\right)-4ac\left(a+2b\right)-2ab\left(a+2b\right)+2c^2\left(a+2b\right)\)

\(B=\left(a+2b\right)\left(bc-4ac-2ab+2c^2\right)\)

\(B=\left(a+2b\right)\left[c\left(2c+b\right)-2a\left(2c+b\right)\right]\)

\(B=\left(a+2b\right)\left(2c+b\right)\left(c-2a\right)\)

\(\left(a^2+b^2+ab\right)^2-a^2b^2-b^2c^2-c^2a^2=\left(a^2+b^2+ab-ab\right)\left(a^2+b^2+2ab\right)-c^2\left(a^2+b^2\right)\)

\(=\left(a^2+b^2\right)\left(a+b\right)^2-c^2\left(a^2+b^2\right)=\left(a^2+b^2\right)\left(a+b-c\right)\left(a+b+c\right)\)