\(a)x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3x^2y-3xy^2+z^3-3xyz=0\)

\(\) \(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(\right.\) \(x^2+2xy+y^2-xz-yz+z^2-3xy)=0\)

Mà \(x+y+z=0\)
\(\Rightarrow0=0\left(đpcm)\right.\)

\(b)\left(x^2y^2+y^2z^2+x^2z^2+2\left.x^2yz+2xy^2z+2xyz^2\right)\right.=x^2y^2+y^2z^2+x^2z^2\)

\(\Leftrightarrow2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)

\(\Leftrightarrow2\left(x+y+z\right)\left(xyz\right)=0\)
Mà \(x+y+z=0\)

\(\Rightarrow0=0\left(đpcm\right)\)

\(c)\) Ta có:\(x+y+z=0\)

\(\Rightarrow\left(x+y+z\right)^2=0\)

\(\Rightarrow x^2+y^2+z^2+2\left(\right.\) \(x^2yz+xy^2z+xyz^2)=0\)

\(\Rightarrow2\left(\right.\) \(xy+yz+xz^{})=-\left(\right.\) \(x^2+y^2+z^2)\)

\(\Rightarrow4\left(\right.\) \(xy+yz+xz)^2=\) \(x^4+y^4+z^4+2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(1\right)\)

Mà ta có: \(\left(xy+yz+xz\right)^2=x^2y^2+y^2z^2+x^2z^2\) (theo câu b)

\(\Leftrightarrow2\left(xy+yz+xz\right)^2=2\left(\right.\) \(x^2y^2+y^2z^2+x^2z^2)\left(2\right)\)

\(\left(1\right)-\left(2\right)\Leftrightarrow2\left(xy+yz+xz\right)^2=x^4+y^4+z^4\left(đpcm\right)\)