Using algebraic identities to prove a number is not prime $$3^{3^n}(3^{3^n}+1)+3^{3^n +1}-1$$
I want the prove that the number is not prime. I used the identity $a^3+b^3+c^3-3abc $. I couldn't simplify to the state where the factors could be observed.

The other approach does show a quite large factor of the number; good to know how to approach this.
The expression $$ 3^{3^n} 3^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1  $$
first becomes
$$  9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
The first two terms really are cubes, since the exponents are divisible by 3. If we look for $a^3 = 3^{3^n},$ it is reasonable to try
$$ a = 3^x  $$
because we want a power of three. Then
$$  a^3 = \left(3^x \right)^3 = 3^{3x}. $$
Then $a^3 = 3^{3^n}$ tells us
$$  3^{3x} = 3^{3^n} $$
so
$$ 3x = 3^n $$ and
$$ x = 3^n 3^{-1} = 3^{n-1},  $$
$$ a = 3^x = 3^{3^{n-1}}  $$
$$  $$
 If we look for $b^3 = 9^{3^n},$ it is reasonable to try
$$ b = 9^y  $$
because we want a power of nine. Then
$$  b^3 = \left(9^y \right)^3 = 9^{3y}. $$
Then $b^3 = 9^{3^n}$ tells us
$$  9^{3y} = 9^{3^n} $$
so
$$ 3y = 3^n $$ and
$$ y = 3^n 3^{-1} = 3^{n-1},  $$
$$ b = 9^x = 9^{3^{n-1}}  $$
$$  $$
We are not yet sure what we want to use for $c.$ Let us examine what we already have for $3ab,$ anticipating the eventual $-3abc.$
$$  a b = 3^{3^{n-1}} 9^{3^{n-1}}  $$
MORE TO COME 11:08 am
We do need to understand $b = a^2.$ Well,
$$ a^2 = \left( 3^{3^{n-1}}\right)^2 = 3^{ 2 \cdot 3^{n-1}} $$ while
$$ b = \left( 3^2 \right)^{3^{n-1}} = 3^{ 2 \cdot 3^{n-1}} \; .  $$
So $b = a^2,$ also $ab = a a^2 = a^3$ so
$$ ab = 3^{3^n} \; .  $$ Next
$$ 3ab = 3 \cdot 3^{3^n} = 3^1 \cdot 3^{3^n} =  \cdot 3^{3^n + 1} \; .$$
This is progress. We now have the overall expression
$$  9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 + 3ab - 1. $$
We can introduce a value of $c$ by demanding $-3abc = 3ab.$ The requires $c=-1.$ We have 
$$ b^3 + a^3 - 3abc - 1.   $$
Finally $c^3 = (-1)^3 = -1,$ and we get
$$  9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 - 3abc - c^3 = a^3 + b^3 + c^3 - 3abc. $$
Perhaps I should say why the new factors are both large. We have
$$a+b+c > b  = 9^{3^{n-1}}$$
It is also worth remembering that
$$ a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2  \right)  $$
which is bigger than $ (b+1)^2 / 2 \; > \; b^2 / 2 \; . $
