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Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)
Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=\left(a^2b-b^2a\right)-\left(a^2c-b^2c\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)\)
\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
=a2b-a2c+b2c-ab2+ac2-bc2
=ab(a-b)-c(a-b)(a+b)+c2(a-b)
=(a-b)(ab-c+c2)
\(A=\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc+abc\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ca\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)
Vậy....
Bạn ơi bạn có thể ghi câu trả lời ra cụ thể giúp mình có được không ạ ?
(a-b)(b-c)(a-c)+(a+b)(c+a)(c-b)+(b+c)(c+a)(b-a)
= ( b - c) [ a2−ab−ac+bc−ac−bc−a2−ab]+(b+c)(c+a)(b−a)a2−ab−ac+bc−ac−bc−a2−ab]+(b+c)(c+a)(b−a)
= -2a.(b + c) (b - c) + (b+c)(c+a)(b-a)
= ( b + c ) ( bc + ab - ac - a2a2 - 2ab + 2ac )
= ( b + c ) ( bc - ab + ac - a2a2 )
= ( b + c ) ( a + b ) ( c - a )
\(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)\)
\(=\left(a-c\right)\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(c+a\right)\left(a+b\right)\left(b-c\right)\)
\(=\left(a-c\right)\left(2ab+2bc\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)\)
\(=2b\left(a-c\right)\left(a+c\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(b-c\right)\right]\)
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)