Đáp án:
a) `a^3+b^3+c^3-3abc=(a+b)^3-3a^2b-3ab^2+c^3-3abc`
`=[(a+b)^3+c^3]-3ab(a+b+c)`
`=(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)`
`=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)`
`=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)`
b) Đặt `x-y=a,y-z=b,z-x=c` thì `a+b+c=0`
Do đó theo câu `a)` ta có `a^3+b^3+c^3-3abc=0=>a^3+b^3+c^3=abc`
`=>(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-z)(z-x)`