Giải:
Vì \(a\in Z^+\)
\(\Rightarrow5^b=a^3+3a^2+5>a+3=5^c\)
\(\Rightarrow5^b>5^c\Rightarrow b>c\)
\(\Rightarrow5^b⋮5^c\)
\(\Rightarrow a^3+3a^2+5⋮a+3\)
\(\Rightarrow a^2\left(a+3\right)+5⋮a+3\)
Mà \(a^2\left(a+3\right)⋮a+3\)
\(\Rightarrow5⋮a+3\)
\(\Rightarrow a+3\inƯ\left(5\right)\)
\(\Rightarrow a+3\in\left\{\pm1;\pm5\right\}\left(1\right)\)
Do \(a\in Z^+\Rightarrow a+3\ge4\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow a+3=5\)
\(\Rightarrow a=5-3\)
\(\Rightarrow a=2\)\((*)\)
Thay \((*)\) vào biểu thức ta có:
\(2^3+3.2^2+5=5^b\Leftrightarrow b=2\)
\(2+3=5^c\Leftrightarrow c=1\)
Vậy: \(\left\{\begin{matrix}a=2\\b=2\\c=1\end{matrix}\right.\)
