Ytukyg

$3^6-3^3 +1$ 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!

It's a special case of: completing the square leads to a difference of squares, i.e.

$$begin{eqnarray}overbrace{3^{large 6}+1}^{rm incomplete}-,3^{large 3}&= &!!!! overbrace{(3^{large 3}+1)^{large 2}}^{rm!!! complete the square!!!}!!!!-color{#c00}{3,3^{large 3}} text{so, factoring this} ittext{ difference of squares}\[.3em]
&!!!=&! (underbrace{3^{large 3}+1, -, color{#c00}{3^{large 2}}}_{Large 19}) (underbrace{3^{large 3}+1 +,color{#c00}{3^2}}_{Large 37})\
end{eqnarray}$$

Generally $ a^{large 6} + b, a^{large 3} + c^{large 2},$ factors if $ color{#c00}{ b= 2c!-!ad^{large 2}}$ for some $d,$ (above is $,a,b,c,d = 3,-1,1,1)$

$$qquad begin{eqnarray}overbrace{a^{large 6}+c^{large 2}}^{rm incomplete}!+b,a^{large 3}&= &!!!! overbrace{(a^{large 3}+c)^{large 2}}^{rm!!! complete the square!!!}!!!!+color{#c00}{overbrace{(b-2c)}^{!!large -d^{Large 2}a},a^{large 3}} text{so, factoring this} ittext{ difference of squares}\[.3em]
&!!!=&! ({a^{large 3}+c, -, color{#c00}{da^{large 2}}}) ({a^{large 3}+c +,color{#c00}{da^{large 2}}})\
end{eqnarray}$$

Remark $ $ Below is another well-known example

$$begin{eqnarray} n^4+4k^4 &,=,& overbrace{(n^2!+2k^2)^2}^{rm!!! complete the square!!!}!!!-!(color{#c00}{2nk})^2 text{so, factoring this} ittext{ difference of squares}\
&,=,& (n^2!+2k^2 -, color{#c00}{2nk}),(n^2!+2k^2+,color{#c00}{2nk})\
&,=,&(underbrace{(n-k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!} + ,k^2) underbrace{((n+k)^2}_{rm!!!!!!!!!!! complete the square!!!!!!!!!!!!!!} +,k^2)\ end{eqnarray}$$

$x^2-x+1$ factorises as $(x-omega)(x+omega^2)$ where $omega^3=-1$.

Here $x=27$ and working modulo $27$ the cubes are $1,8,0,10,17,0,19,26, 0, 1, 8 dots$ so $8^3equiv -1$, and we can take $omega = 8, omega^2=64equiv 10$ and obtain the factorisation $$(27-8)(27+10)=19times 37$$

Since we have a cube root involved and the modulus is a power of $3$ there are some curiosities about the factorisation, but it checks out all the same.

Numbers having the form $n^2-n+1$ can never have prime factors congruent to $2$ modulo $3$. Since $3$ is obviously out and any factorization must include a prime factor less than or equal to the square root if the number is composite, only factors $7,13,19$ need further investigation. The factorization $19×37$ is tgen fairly easy to find.