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

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Answer 1 · 2015-11-20T14:37:08

#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)#

First use the sum of cubes identity:

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

with #A=a^2# and #B=5# to find:

#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 #a^2+5# nor #a^4-5a^2+25# have linear factors with Real coefficients, but #a^4-5a^2+25# will have quadratic factors with Real coefficients.

Consider:

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

If we let #k = sqrt(15)# then:

#(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)#