+) With n = 1 then 

\(3^{2n+3}+40n-27=256⋮64\)

+) Suppose that \(\left(3^{2n+3}+40n-27\right)⋮64\)

And we will prove \(\left(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27\right)⋮64\) , too 

Also \(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27=3^{2n+3}.9+40n+40-27\)

                                                         \(=3^{2n+3}.9+9.40n-27.9-8.40n+27.8+40\)

                                                         \(=9\left(3^{2n+3}+40n-27\right)-320n+256\)

                                                         \(=9\left(....\right)-64\left(5n+4\right)⋮64\)

So \(\left(3^{2\left(n+1\right)+3}+40\left(n+1\right)-27\right)⋮64\)  is true 

Conclude \(\left(3^{2n+3}+40n-27\right)⋮64\)        \(\forall n\in N\)