Ta có:
\(\dfrac{\overline{ab}}{b}=\dfrac{\overline{bc}}{c}=\dfrac{\overline{ca}}{a}\)
\(\Rightarrow\dfrac{10a}{b}+\dfrac{b}{b}=\dfrac{10b}{c}+\dfrac{c}{c}=\dfrac{10c}{a}+\dfrac{a}{a}\)
\(\Rightarrow\dfrac{10a}{b}+1=\dfrac{10b}{c}+1=\dfrac{10c}{a}+1\)
\(\Rightarrow\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{10c}{a}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{10a}{b}=\dfrac{10b}{c}=\dfrac{10c}{a}=\dfrac{10a+10b+10c}{b+c+a}=\dfrac{10\left(a+b+c\right)}{a+b+c}=10\)
\(\Rightarrow\left\{{}\begin{matrix}10a=10b\\10b=10c\\10c=10a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow\left(\overline{abc}\right)^{123}=\left(\overline{aaa}\right)^{123}\)(1)
\(\Rightarrow c=111^{123}.a^{40}.a^{41}.a^{42}=111^{123}.a^{123}=\left(111.a\right)^{123}=\left(\overline{aaa}\right)^{123}\)(2)
Từ (1) và (2) suy ra: \(\left(\overline{abc}\right)^{123}=111^{123}.a^{40}.b^{41}.c^{42}\)
