Factoring $3^6-3^3 +1$ $3^6-3^3 +1$ factors?, 37 and 19, but how to do it using factoring, $3^3(3^3-1)+1$, can't somehow put the 1 inside  

Viewing this as a quadratic in $3^3$ leads you to try to factorise $x^2-x+1$ or $x^6-x^3+1$, neither of which splits into anything useful. However, you can pull out a factor of $9$ from $3^6$ and a factor of $3$ from $3^2$ to obtain
$$3^6-3^3+1 = 9 \cdot 3^4 - 3 \cdot 3^2 + 1$$
And the quartic $9x^4-3x^2+1$ does factorise into two quadratic factors:
$$9x^4-3x^2+1 = (3x^2+3x+1)(3x^2-3x+1)$$
Now set $x=3$ and voilà!
