Giải:
Do \(a\in Z^+\)
\(\Rightarrow5^b=a^3+3a^2+5>a+3=5^c\)
\(\Rightarrow5^b>5^c\Rightarrow b>c\Rightarrow5^b⋮5^c\)
\(\Rightarrow a^3+3a^2+5⋮a+3\)
\(\Rightarrow a^2\left(a+3\right)+5⋮a+3\)
Mà \(a^2\left(a+3\right)⋮a+3\)
\(\Rightarrow5⋮a+3\Rightarrow a+3\inƯ\left(5\right)\)
\(\Rightarrow a+3=\left\{\pm1;\pm5\right\}\left(1\right)\)
Do \(a\in Z^+\Rightarrow a+3\ge4\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow a+3=5\Rightarrow a=2\)
\(\Rightarrow2^3+3.2^2+5=5^b=5^5\Leftrightarrow b=5\)
\(\Rightarrow2+3=5^c=5^1\Leftrightarrow c=1\)
Vậy: \(\left\{{}\begin{matrix}a=2\\b=2\\c=1\end{matrix}\right.\)
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