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 \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)

\(=\left(ab^3-a^3b\right)+\left(bc^3-ac^3\right)+\left(a^3c-b^3c\right)\)

\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)

\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)

Ta có: $VT={}^{}-{}^{}-{}^{}-{}^{}$

$=\left[{}^{}-{}^{}\right]-\left({}^{}+{}^{}\right)$

$=\left(b+c\right)\left[{}^{}+\left(a+b+c\right)a+{}^{}\right]-\left(b+c\right)\left({}^{}-bc+{}^{}\right)$

$=\left(b+c\right)\left(3{}^{}+3ab+3bc+3ca\right)$

$=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]$

$=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP$ (Đpcm)

Thật ra mình làm theo đề thấy nó đáng ra phải là chứng minh chứ ko phải phân tích . chúc học tốt!

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=(a−b)(b−c)[(a−c)(a+c)+b(a−c)]

=(a−b)(b−c)(a−c)(a+b+c)