How do you factor `a^6 + 7a^3 + 6`?

Call `x = a^3`. Factor the quadratic equation:
`f(x) = x^2 + 7x + 6.`
Since a - b + c = 0, use shortcut. One factor is (x + 1) and the other is `(x + c/a) = (x + 6)`. We get:
`f(x) = (x + 1)(x + 6)`
Replace x by a^3.
`f(a) = (a^3 + 1)(a^3 + 6)`
Factor `(a^3 + 1)` and `(a^3 + 6)` by applying the algebraic identity:
`a^3 + b^3 = (a + b)(a^2 - ab + b^2)`.
Finally,
`f(a) = (a + 1)(a^2 - a + 1)(a^3 + 6)`

First treat this as a quadratic in `a^3`, noting that `1+6=7` and `1*6=6`.

So:

`a^6+7a^3+6 = (a^3)^2+7(a^3)+6 = (a^3+1)(a^3+6)`

Note that both `a^3` and `1=1^3` are perfect cubes, so we can use the sum of cubes identity:

`A^3+B^3 = (A+B)(A^2-AB+B^2)`

with `A=a` and `B=b` to find:

`(a^3+1) = (a^3+1^3) = (a+1)(a^2-a+1)`

The factor `(a^3+6)` is not quite so nice, but it can still be treated as a sum of cubes, using `A=a` and `B=root(3)(6)` as follows:

`(a^3+6)`

`= (a^3+(root(3)(6))^3)`

`= (a+root(3)(6))(a-root(3)(6)a+(root(3)(6))^2)`

`= (a+root(3)(6))(a-root(3)(6)a+root(3)(6^2))`

`= (a+root(3)(6))(a-root(3)(6)a+root(3)(36))`

Putting this together:

`a^6+7a^3+6=(a+1)(a^2-a+1)(a+root(3)(6))(a-root(3)(6)a+root(3)(36))`

This is as far as we can go with Real numbers.