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\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)
\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)
\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)
\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
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 \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)
\(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a+a-c\right)+c^2a^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+b^2c^2\left(b-a\right)+b^2c^2\left(a-c\right)+c^2a^2\left(c-a\right)\)
\(=b^2\left(a-b\right)\left(a^2-c^2\right)+c^2\left(c-a\right)\left(a^2-b^2\right)\)
\(=b^2\left(a-b\right)\left(a-c\right)\left(a+c\right)+c^2\left(c-a\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[-b^2\left(a+c\right)+c^2\left(a+b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(-ab^2-b^2c+ac^2+bc^2\right)\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c^2-b^2\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left[a\left(c-b\right)\left(c+b\right)+bc\left(c-b\right)\right]\)
\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(ab+bc+ca\right)\)
Lời giải :
\(B=2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)
\(B=2b^2c+4bc^2+2ac^2-4a^2c-2ab\left(a+2b\right)-7abc\)
\(B=abc+2b^2c-4a^2c-8abc-2ab\left(a+2b\right)+2ac^2+4bc^2\)
\(B=bc\left(a+2b\right)-4ac\left(a+2b\right)-2ab\left(a+2b\right)+2c^2\left(a+2b\right)\)
\(B=\left(a+2b\right)\left(bc-4ac-2ab+2c^2\right)\)
\(B=\left(a+2b\right)\left[c\left(2c+b\right)-2a\left(2c+b\right)\right]\)
\(B=\left(a+2b\right)\left(2c+b\right)\left(c-2a\right)\)
2bc(b + 2c) + 2ac(c - 2a) - 2ab(a + 2b) - 7abc
= 2b2c + 4bc2 + 2ac2 - 4a2c - 2ab(a + 2b) - 7abc
= 2b2c + abc + 4bc2 + 2ac2 - 4a2c - 8abc - 2ab(a + 2b)
= bc(2b + a) + 2c2(2b + a) - 4ac(a + 2b) - 2ab(a + 2b)
= (a + 2b)(bc + 2c2 - 4ac - 2ab)
= (a + 2b)[c(b + 2c) - 2a(2c + b)]
= (a + 2b)(b + 2c)(c - 2a)