Lời giải:

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

\(=a(b^3-c^3)-b[(b^3-c^3)+(a^3-b^3)]+c(a^3-b^3)\)

\(=(b^3-c^3)(a-b)-(a^3-b^3)(b-c)\)

\(=(b-c)(a-b)(b^2+bc+c^2)-(a-b)(b-c)(a^2+ab+b^2)\)

\(=(a-b)(b-c)(b^2+bc+c^2-a^2-ab-b^2)\)

\(=(a-b)(b-c)[(c-a)(c+a)+b(c-a)]\)

\(=(a-b)(b-c)(c-a)(c+a+b)\)

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!

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

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

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