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

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

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

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

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

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

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

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

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

chưa tối giản :v

M = (a + b + c)³ - a³ - b³ - c³

= (a + b)³ + c³ + 3(a + b)²c + 3(a + b)c² - a³ - b³ - c³

= a³ + b³ + c³ + 3a²b + 3ab² + 3(a + b)c(a + b + c) - a³ - b³ - c³

= 3ab (a + b) + 3c(a + b)(a + b + c)

= 3(a + b)[ab + c(a + b + c)]

= 3(a + b)(ab + bc + ac + c²)

= 3(a + b)[b(a + c) + c(a + c)]

= 3(a + b)(b + c)(c + a)

N = a³ + b³ + c³ - 3abc

= (a + b)³ + c³ - 3ab(a + b) - 3abc

= (a + b + c)³ - 3(a + b)c(a + b + c) - 3ab(a + b + c)

= (a + b + c)[(a + b + c)² - 3(a + b)c - 3ab]

= (a + b + c)(a² + b² + c² + 2ab + 2bc + 2ca - 2ac - 3bc - 3ab)

= (a + b + c)(a² + b² + c² - ab - bc - ca)

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

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

Cảm ơn Nguyễn Đức Anh :))

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

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

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Ta có: $VT={\left(a+b+c\right)}^3-a^3-b^3-c^3$

$=\left[{\left(a+b+c\right)}^3-a^3\right]-\left(b^3+c^3\right)$

$=\left(b+c\right)\left[{\left(a+b+c\right)}^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)$

$=($