Ta có x+y+z=0<=>(x+y+z)²=0<=>x² +y²+z²+2(xy+yz+zx)=0
Vì x² + y² + z² =4 nên 4+2(xy+yz+zx)=0<=>xy+yz+zx = \(-2\)
⇒(xy+yz+zx)² = 4<=>x²y²+y²z² +z²x²+2xyz(x+y+z)=4
Vì x+y+z=0 nên x²y+y²z+z²x²=4.
Ta có:
x² + y² + z² = 4 <=> (x² + y² + z²)² = 16 <=> x⁴ + y⁴+z⁴+2(x²y² + y²z² +z²x²) = 16
Mà  x²y²+y²z²+z²x² =4nên x⁴ +y⁴+z⁴ =8.

a) 6x²-3xy=3x(2x-y) 
b) x²-y²-6x+9= x²-6x+9-y² = (x-3)²-y²=(x-3-y)(x-3+y)
c) x²+5x-6= x²-x+6x-6=(x-1)(x+6)