x³+ y³= z³
<=> (x+y)³- 3xy (x+y)- z³=0
<=> (x+y-z)(x²+ 2xy+y² + xz+ yz+ z²)- 3xy (x+y)+3xyz= 3xyz
<=> (x+y-z)(x²+ 2xy+y² + xz+ yz+ z²)- 3xy (x+y-z)= 3xyz
<=> (x+y-z)(x²-xy+y² + xz+ yz+ z²)= 3xyz
<=> (x+y-z)(2x²-2xy+2y² + 2xz+ 2yz+ 2z²)= 6xyz
<=> (x+y-z)[(x-y)²+ (y+z)²+ (z+x)²]= 6xyz
vì (x-y)²+ (y+z)²+ (z+x)² ≥ 0 => (x+y-z)[(x-y)²+ (y+z)²+ (z+x)²] ≥ 0
=> 6xyz ≥ 0
<=> xyz ≥ 0
Dấu "=" xảy ra <=> x=0 hoặc y=0 hoặc z=0
Mà x,y,z ∈ N*
=> pt vô nghiệm
