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

##### 2 Answers

#### Explanation:

Call

Since a - b + c = 0, use shortcut. One factor is (x + 1) and the other is

Replace x by a^3.

Factor

Finally,

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

#### Explanation:

First treat this as a quadratic in

So:

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

Note that both

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

with

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

The factor

#(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.