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a) \(A=a^3-b^3-c^3-3abc\)
\(=\left(a-b\right)^3+3ab\left(a-b\right)-c^3-3abc\)
\(=\left(a-b-c\right)\left[\left(a-b\right)^2+c\left(a-b\right)+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)\)
\(=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
b) \(B=a^2b^2\left(a-b\right)-c^2b^2\left(c-b\right)+a^2c^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)+a^2c^2\left(c-a\right)\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left[\left(a-b\right)+\left(b-c\right)\right]\)
\(=a^2b^2\left(a-b\right)+c^2b^2\left(b-c\right)-a^2c^2\left(a-b\right)-a^2c^2\left(b-c\right)\)
\(=a^2\left(a-b\right)\left(b^2-c^2\right)+c^2\left(b-c\right)\left(b^2-a^2\right)\)
\(=a^2\left(a-b\right)\left(b-c\right)\left(b+c\right)+c^2\left(b-c\right)\left(b-a\right)\left(b+a\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2b+a^2c-bc^2-ac^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)
1.
\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)
Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)
Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)
Từ đó ta được đpcm
\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)
\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)
\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)
\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)
\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)
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 \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)
\(B=x^3+3x^2-4\)
\(B=x^3-x^2+4x^2-4\)
\(B=x^2\left(x-1\right)+4\left(x^2-1\right)\)
\(B=x^2\left(x-1\right)+4\left(x+1\right)\left(x-1\right)\)
\(B=\left(x-1\right)\left(x^2+4x+4\right)\)
\(B=\left(x-1\right)\left(x+2\right)^2\)