Question

[Quay Cuồng]

cho \(\overline{abc}\)\(\in Z\)thỏa mãn (a-b)(b-c)(c-a)=a+b+c .CMR a+b+c\(⋮27\)

Answer 1

[Nguyen Viet Bac]

Ta có :

\(\left(a-b\right)\left(b-c\right)\left(c-a\right)=bc^2+b^2a+a^2c-b^2c-ac^2-a^2b\)

=> \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=c^2\left(b-a\right)+b^2\left(a-c\right)+a^2\left(c-b\right)\)

=> \(c^2\left(b-a\right)+b^2\left(a-c\right)+a^2\left(c-b\right)=a+b+c\)

=> \(a\left(ac-ba-1\right)+b\left(ba-ba-1\right)+c\left(cb-ca-1\right)=0\)

TH1:

\(\hept{\begin{cases}a=0\\b=0\\c=0\end{cases}}\)

TH2:

\(\hept{\begin{cases}\left(ac-ab-1\right)=0\\\left(ab-bc-1\right)=0\\\left(bc-ac-1\right)=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(ac-ab\right)=1\\\left(ab-bc\right)=1\\\left(bc-ac\right)=1\end{cases}}\)\(\Rightarrow\)\(\left(ac-ab\right)+\left(ab-bc\right)+\left(bc-ac\right)=3\)

=> 0 = 3 (loại)

Vậy a = b = c = 0