Two factoring problems that I can't figure out I have two problems I can't seem to factor.


*

*$8a^3+27b^3+2a+3b$


I first tried it by grouping. I grouped the first two terms together and used sum of two cubes to get
$$(2a+3b)(4a^2-6ab+9b^2)$$ and then I added the leftover $2a+3b$ to get:
$$(2a+3b)(4a^2-6ab+9b^2) + 2a+3b$$
But the answer is $(2a+3b)(4a^2-6ab+9b^2 + 1)$ and I'm not quite sure how you got the $+1$ from the $2a+3b$.


*$x^3+3x^2+3x+1$


The answer is $(x+1)^3$ but I don't know how to get there. I tried factoring by grouping, but that wasn't right, and I'm stuck as to how to solve it.
Thanks!

.For the first question, you were very close to the answer. In fact, when you write $(2a+3b)(4a^2+9b^2-6ab) + 2a+3b$, all you need to notice is that the $2a+3b$ at the end can also be taken into the bracket, using the distributive law:
$$
(2a+3b)(4a^2+9b^2-6ab) + 2a+3b = (2a+3b)(4a^2+9b^2-6ab) + 1(2a+3b) \\ = (2a+3b)(4a^2+9b^2-6ab + 1)
$$
Using $xy+ xz = x(y+z)$, where $x=2a+3b, y = 4a^2+9b^2-6ab,z=1$.
Hence the answer follows.
For the second one, yes it is tricky, but here is the catch,  or rather a heuristic in these kind of problems:

Suppose you are asked to factorize some complicated cubic, say $x^3+ax^2+bx+c$, or any other complicated expression. Then, most likely, the roots are integers, which are factors of the constant in the expression.

For example, here $x^3+3x^2+3x+1$ is given, and the constant is $1$. Hence, the roots, if they are integers, are all factors of $1$. Then, they can only be $-1$ or $1$. You can use the remainder theorem to check which of these is a factor, by substituting $-1$ and $1$ in the expression and checking when the value is zero. Turns out for $-1$ that the expression is $0$, so $(x-(-1)) = (x+1)$ is a factor of your expression. Then, go on and divide by $1$, you will get $x^2+2x+1$, which is $(x+1)^2$. So collecting everything, we have $x^3+3x^2+3x+1=(x+1)^3$.
It's good to do one more example. Suppose someone asks you to factor $x^3-x^2-4x+4$. Using the heuristic, you know that the roots must divide $4$. Hence, the roots can only be $\pm 1,\pm 2,\pm4$. Use remainder theorem, and check that substituting $1$ gives $0$. Hence $x-1$ is a factor, and on dividing you get $x^2-4$, which is $(x-2)(x+2)$. Hence the complete answer is $(x-1)(x+2)(x-2)$.
