a: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{yz+xz+xy}{xyz}=0\)

=>xy+xz+yz=0

Đặt a=xy; b=xz; c=yz

=>a+b+c=0

=>\(\left(a+b+c\right)^2=0\)

=>\(a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left\lbrack a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)\right\rbrack\)

=>\(\left(a^2+b^2+c^2\right)^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)

Ta có: \(\left(a^2+b^2+c^2\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

=>\(2\left(a^2b^2+b^2c^2+a^2c^2\right)=\left(a^2+b^2+c^2\right)^2-\left(a^4+b^4+c^4\right)\)

=>\(4\left(a^2b^2+b^2c^2+a^2c^2\right)=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

DO đó, ta có: \(\left(a^2+b^2+c^2\right)^2=2\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)\)

=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

=>\(2\left(x^4y^4+y^4z^4+x^4z^4\right)=\left(x^2y^2+z^2y^2+x^2z^2\right)^2\)

b:

x+y+z=0

=>x+y=-z

\(x^3+y^3+z^3-3xyz\)

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

\(=\left(-z\right)^3-3xy\cdot\left(-z\right)-3xyz+z^3=z^3+3xyz-3xyz-z^3=0\)

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

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

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

\(=\left[x^2\left(y-z\right)-y^2\left(y-z\right)\right]+\left[y^2\left(y-x\right)-z^2\left(y-x\right)\right]\)

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

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

\(=\left(x-y\right)\left(y-z\right)\left[\left(x+y\right)-\left(y+z\right)\right]\)

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

Thanks you!!!

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

\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)

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

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

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

           ~ Chúc bạn học tốt~

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

\(=x^3+3x^2yz+3xy^2z+3xyz^2+y^3+z^3-x^3-y^3-z^3\)

\(=3x^2yz+3xy^2z+3xyz^2\)

\(=3xyz\left(x+y+z\right)\)

bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

1). x²y²(y-x)+y²z²(z-y)-z²x²(z-x)

2)xyz-(xy+yz+xz)+(x+y+z)-1

3)yz(y+z)+xz(z-x)-xy(x+y)

5)y(x-2z)²+8xyz+x(y-2z)²-2z(x+y)²

6)8x³(y+z)-y³(z+2x)-z³(2x-y)

7) (x2+y2)³+(z2-x2)³-(y2+z2)³

\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)=x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)\)

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

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

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

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

\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

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