Em thử nha, sai thì thôi
Đặt a - b = x; b-c = y suy ra c - a = -(x+y)
Ta có: \(a^2b^2x+b^2c^2y-c^2a^2\left(x+y\right)\)
\(=a^2x\left(b^2-c^2\right)+c^2y\left(b^2-a^2\right)\)
\(=a^2\left(b+c\right)\left(a-b\right)\left(b-c\right)-c^2\left(a+b\right)\left(a-b\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2b-bc^2+a^2c-c^2a\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(a-c\right)\left(a+c\right)+ac\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)
Vậy...
\(a^2b^2(a-b)+b^2c^2(b-c)+c^2a^2(c-a)\)
\(= a^2b^2(a-b)-b^2c^2\left[\left(a-b\right)+\left(c-a\right)\right]+c^2a^2(c-a)\)\(= a^2b^2(a-b)-b^2c^2(a-b)-b^2c^2(c-a)+c^2a^2(c-a)\\ = b^2(a-b)(a^2-c^2)+c^2(c-a)(a^2-b^2)\\ = b^2(a-b)(a-c)(a+c)-c^2(a-c)(a-b)(a+b)\)
\(= (a-c)(a-b)\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)
\(= (a-c)(a-b)(b^2a+b^2c-c^2a-c^2b)\)
\(= (a-c)(a-b)\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)
\(= (a-c)(a-b)\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)\(= (a-c)(a-b)(b-c)\left[a\left(b+c\right)+bc\right]\\ = (a-c)(a-b)(b-c)(ab+ac+bc)\)
