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Đặt \(f=a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-a-b\right)\)
\(=3abc+a^3+b^3+c^3-a^2b-b^2a-a^2c-b^2c-c^2a-c^2b\)
\(=a^2\left(a-b\right)+b^2\left(b-a\right)+c\left[2ab-a^2-b^2+c\left(c^2-bc-ac+ab\right)\right]\)
\(=\left(a-b\right)\left(a^2-b^2\right)-c\left(a-b\right)^2+c\left(c-a\right)\left(c-b\right)\)
\(=\left(a-b\right)^2\left(a+b+c\right)+c\left(b-c\right)\left(a-c\right)\)
\(\Rightarrow BT=\left(a-b\right)^2\left(a+b+c\right)+c\left(b-c\right)\left(a-c\right)-c\left(b-c\right)\left(a-c\right)\)
\(=\left(a+b\right)^2\left(a+b+c\right)\)
a) \(\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3c\left(a+b\right)\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
c) \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)