Đặt: $3^n-8=a^3(a\in N)$
$\to 3^n=a^3+8\\\to 3^n=(a+2)(a^2-2a+4)\\\to a+2,a^2-2a+4| 3^n$
Đặt: $\begin{cases} a+2=3^x(1)\\a^2-2a+4=3^y (2)\end{cases}(x,y\in N,x+y=n)$
$(1)\to a=3^x-2\\(2)\to (3^x-2)^2 - 2(3^x-2)+4=3^y\\\to 3^{2x} - 6.3^x +12=3^y$
Nếu $x=0\to a+2=1\to a=-1$(Loại).
Nếu $x=1\to a+2=3\to a=1\to n=2$(Nhận)
Nếu $x\ge 2$
$\to 3^{2x}-6.3^x≡0\pmod{9}\\\to 3^{2x}-6.3^x+12≡3\pmod{9}$
Mặt khác: $3^{2x}-6.3^x +12\ge 39 > 3^3\to 3^y>3^3\to y>3$
$\to 3^y≡0\pmod{9}$(Vô lí).
Vậy $n=2$
