Question

(x-y)³+(y-z)³+(z-x)³

Answer 1

Đặt a=x-y, b=y-x, c=z-x

=>a+b+c=0

Ta có : a+b+c=0

=>\(\left(a+b+c\right)^3=0\)

=>\(a^3+b^3+c^3+3a^2b+3a^2c+3b^2a+3b^2c+3c^2a+3c^2b+6abc=0\)=>\(a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)-3abc=0\)

mà a+b+c=0

=>\(a^3+b^3+c^3=3abc\)

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