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Đặt \(x+y-z=a;x-y+z=b;y+z-x=c\)
Ta có:\(A=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(A=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(A=\left(a+b\right)^3+3\left(a+b\right)\cdot c\cdot\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)+c^3-a^3-b^3-c^3\)
\(A=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(A=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Hay \(A=3\cdot2x\cdot2y\cdot2z\)
\(A=24xyz\)
a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3x^2y+3x^2z+3y^2z+3xy^2+3xz^2+3yz^2+6xyz-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3x^2z+3xz^2+3y^2z+3yz^2+6xyz\)
\(=3xy\left(x+y\right)+3xz\left(x+z\right)+3yz\left(y+z\right)+6xyz\)
\(=3\left[xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+2xyz\right]\)
\(=3\left[xy\left(x+y\right)+x^2z+xz^2+y^2z+yz^2+2xyz\right]\)
\(=3\left[xy\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+yz\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]+\left(z-x\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-2y+z\right)-\left(x-y\right)\left(y-2z+x\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-z\right)\left(x-y\right)\left(3z-3y\right)\)
\(=3\left(x-z\right)\left(x-y\right)\left(z-y\right)\)
Đặt y-z=-[(x-y)+(z-x)]
Thay vào rồi cm nha bạn