`a^3+3a^2+5=5^b`
`⇒a^2(a+3)+5=5^b`
`⇒a^2 5^c+5=5^b`
`⇒5(5^(c-1)a^2+1)=5^b`
`⇒5^(c-1)a^2+1=5^(b-1)`
`⇒`\(\left[ \begin{array}{l}c-1=0\\b-1=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}c=1\\b=1\end{array} \right.\)
+) TH1: `b=1`
`5^(c-1)a^2+1=5^(1-1)`
`⇒5^(c-1)a^2+1=1`
`⇒5^(c-1)a^2=0 (`loại vì `a,b,c\ne0)`
+) TH2: `c=1`
`5^(1-1)a^2+1=5^(b-1)`
`⇒a^2+1=5^(b-1)`
`⇒a=2` và `b=2`
Vậy `a=2 , b=2 , c=1 `