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

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

\(=\left(x+y\right)^3+3z\left(x+y\right)^2+3\left(x+y\right)\cdot z^2+z^3-\left(x+y\right)^3+3z\left(x+y\right)^2-3\left(x+y\right)\cdot z^2+z^3\)

\(=6z\left(x+y\right)^2+2z^3\)

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

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

\(=\left(x-y\right)^3+3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2+z^3+\left(x-y\right)^3-3\left(x-y\right)^2\cdot z+3\left(x-y\right)\cdot z^2-z^3\)

\(=2\left(x-y\right)^3+6\left(x-y\right)\cdot z^2\)

=>\(A=6z\left(x+y\right)^2+2z^3+2\left(x-y\right)^3+6z^2\left(x-y\right)\)

a) Các biểu thức: \(\dfrac{1}{5}x{y^2}{z^3}; - \dfrac{3}{2}{x^4}{\rm{yx}}{{\rm{z}}^2}\) là đơn thức

b) Các biểu thức: \(2 - x + y; - 5{{\rm{x}}^2}y{z^3} + \dfrac{1}{3}x{y^2}z + x + 1\) là đa thức

Ta sử dụng ẩn phụ:

\(\hept{\begin{cases}a=x+y-z\\b=y+z-x\\c=x+z-y\end{cases}}\)=> x+y+z=a+b+c

Khi đó :

A= (x+y+z)^3-(x+y-z)^3-(-x+y+z)^3-(x-y+z)^3=(a+b+c)^3+a^3+b^3+c^3=3(a+b)(b+c)(c+a)=3*2y*2z*2x=24xyz