Question

\(\text{Chứng minh rằng:}\)

\(\text{Nếu}\)\(x^3+y^3+z^3-3xyz=0.\)\(\text{Thì}\)\(x+y+z=0\)\(\text{hoặc}\)\(x=y=z\)
 

Answer 1

x^3 + y^3 + z^3 - 3xyz = (x+y)^3 + z^3 - 3x^2y - 3xy^2 - 3xyz 
= (x+y)^3 + z^3 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)^2.z - 3(x+y)z^2 - 3xy(x + y + z) 
= (x+y+z)^3 - 3(x+y)z(x+ y + z) - 3xy(x + y + z) 
=(x+y+z)[(x+y+z)^2 - 3(x+y)z - 3xy] 

=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

=1/2(x+y+z)(x^2-2xy+y^2+y^2-2yz+z^2+x^2-2xz+z^2)

=1/2(x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2]

mà x^3 + y^3 + z^3 - 3xyz=0

<=> x+y+z=0

Vậy ...

Chúc bạn học tốt .

hoặc (x-y)^2+(y-z)^2+(x-z)^2 =0 mà (x-y)^2,(y-z)^2,(x-z)^2 >=0 mọi x,y,z

=> x-y=y-z=x-z=0 => x=y=z