$\begin{align} \text{Ta có : } \;
(A+B+C)^3 & = (A+B+C)(A+B+C)(A+B+C) \\
& = (A^2+AB+AC+B^2+AB+BC+C^2+AC+BC)(A+B+C) \\
&= (A^2+2AB+B^2+2BC+C^2+2AC)(A+B+C) \\
&= A^3+2A^2B+AB^2+2ABC+AC^2+2A^2C+BA^2+2AB^2+B^3+2B^2C+BC^2+2ABC+A^2C+2ABC+B^2C+2BC^2+C^3+2AC^2 \\
&= A^3+B^3+C^3+3A^2B+3AB^2+6ABC+3AC^2+3A^2C+3B^2C+3BC^2 \\
&= A^3+B^3+C^3+3(A^2B+AB^2+A^2C+AC^2+B^2C+BC^2+2ABC) \\
&= A^3+B^3+C^3+3(A^2B+AB^2+ABC+A^2C+AC^2+ABC+B^2C+BC^2) \\
&= A^3+B^3+C^3+3[AB(A+B+C)+CA(A+B+C)+BC(B+C)] \\
&= A^3+B^3+C^3+3[A(B+C)(A+B+C)+BC(B+C)] \\
&= A^3+B^3+C^3+3[(B+C)(A^2+AB+AC)+BC(B+C)] \\
&= A^3+B^3+C^3+3(B+C)(A^2+AB+AC+BC) \\
&= A^3+B^3+C^3+3(B+C)[A(A+B)+C(A+B)] \\
&= A^3+B^3+C^3+3(B+C)(A+C)(A+B) \\
\implies (A+B+C)^3 &= A^3+B^3+C^3+3(B+C)(A+C)(A+B) \\
\implies (A+B+C)^3& -A^3-B^3-C^3 = 3(A+B)(B+C)(A+C) \; (1)
\end{align} $
$(x+y+z)^3-(y+z-x)^3-(x+z-y)^3-(x+y-z)^3$
Đặt $a = y+z-x,b = x+z-y,c=x+y-z$
Ta có : $a+b+c =y+z-x+x+z-y+x+y-z=x+y+z$
$ \begin{align}
\text{Áp dụng (1) :}\; (x+y+z)^3-(y+z-x)^3-(x+z-y)^3-(x+y-z)^3 &\iff (a+b+c)^3-a^3-b^3-c^3 \\
&= 3(a+b)(b+c)(a+c) \\
&\iff 3(y+z-x+x+z-y)(x+z-y+x+y-z)(y+z-x+x+y-z) \\
&= 3.2z.2x.2y \\
&=24xyz
\end{align}$
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