# How do you factor #a^6+125#?

##### 1 Answer

Nov 20, 2015

#### Answer:

#a^6+125 = (a^2+5)(a^4-5a^2+25)#

#=(a^2+5)(a^2+sqrt(15)a+5)(a^2-sqrt(15)a+5)#

#### Explanation:

First use the sum of cubes identity:

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

with

#a^6+125 = (a^2)^3+5^3 = (a^2 + 5)((a^2)^2 - (a^2)(5) + 5^2)#

#= (a^2+5)(a^4-5a^2+25)#

Neither

Consider:

#(a^2+ka+5)(a^2-ka+5) = a^4+(10-k^2)a^2+25#

If we let

#(a^2+ka+5)(a^2-ka+5) = a^4+(10-15)a^2+25 = a^4-5a^2+25#

So:

#a^4-5a^2+25 = (a^2+sqrt(15)a+5)(a^2-sqrt(15)a+5)#