Find all the prime factors of $1000027$ 
Find all the prime factors of $1,000,027$.

I got all the factors by testing every number from $1$ to $103$:P, but when I try to do it using algebra, I get stuck.
My work:

$$1000027$$
$$=(100+3)(100^2-3\cdot100+3^2)$$

How do I simplify further?
 A: I think that the only important step is the first one.
If you want to find the prime factors of $73\times 103$ it is going to be tough, because you have to try all primes up to $73$.
On the other hand, once you find the factor $103$, factoring $7\times 19\times 79$ is easy by brute force, because the factors $7$ and $19$ are found really fast, and then proving that $79$ is prime is done quickly also.
A: First of all we want use $$a^3-b^3=(a+b)(a^2-ab+b^2)$$
we find $1000027=103 \cdot 9709$
then we want to use $a^2-b^2=(a+b)(a-b)$. With a clever observation we see that $$9709=10609-900=(103+30)(103-30)=133\cdot 73$$
that can be factored with ease!
A: To begin with note that
$1000027 = 100^3 + 3^3 = (100+3)(100^2−3⋅100+32) = 103 \cdot 9709 = 103 \cdot (1001 \cdot 9 + 700) = 103 \cdot 7 \cdot (13 \cdot 11 \cdot 9+100) = 103 \cdot 7 \cdot 1387$
where we have used the well-known fact that $1001 = 7 \cdot 11 \cdot 13$
Then I guess $1387$ is small enough to factor by testing cases. Using well-known divisibility rules we can see rather quickly that $2,3,5,7$ and $11$ are not prime factors of $1387$. Also $13$ can be dismissed since $13 \nmid 87$. Also $1387 = 1700-313 = 1700 -340 + 27$, and hence $17 \nmid 1387$. Then testing $19$ we see that $1387 = 1900 - 513 = 1900 - 570 + 57$, and hence $19 \mid 1387$.
Then it is straightforward to check that $1000027 = 7 \cdot 19 \cdot 73 \cdot 103$ 
A: Use the fact that $73 \times 137=10001 = 10^4+1$.
Now,  mark off the number in groups of four digits starting from the right, and add the four-digit groups together with alternating signs.
Applying the above rule, $1000027$ in groups of $4$ is $\underbrace{0100}$ $\underbrace{0027}$. Adding the groups with alternate signs gives $73$. Therefore, $1000027$ is divisble by $73$ and gives $13699$ as quotient.
For $13699$ apply the divisbility test by $7$ by marking of the digits in groups of $3s$ by starting from the right and adding together with alternate signs. Therefore, adding the groups $\underbrace{013}$ $\underbrace{699}$ with alternate signs gives $686$ which is divsible by $7$ and hence $13699$ is disvisble by $7$ 
$13699$ when divided by $7$ gives $1957$.
Now, $1957$ is $1900+57$ and hence is divisible by $19$ giving the quotient as $103$.
Combining them all gives $1000027 = 7\times19\times73\times103$
A: Go with the sum of cubes and factor out the $103$ thing (which you already did), then notice that
$$100^2-3\cdot100+3^2= \\
=100^2+2\cdot3\cdot100+3^2-3\cdot3\cdot100=\\
=(100+3)^2-30^2$$
then factor it as a difference of squares. Then you'll only have to factor $133=7\cdot19$ by hand and verify that everything else is prime.
Come to think of it, this might be an example of Aurifeuillean factorization. Pretty much everybody knows that $x^4+4=(x^2+2x+2)(x^2-2x+2)$. Now we used (rediscovered, if you'd like) a more complicated thing of the same sort:
$$x^6+27=(x^2+3)(x^2+3x+3)(x^2-3x+3)$$
