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.