How do you factor completely #a^3 - 8#?

1 Answer
George C.
Jan 4, 2016

#a^3-8 = (a-2)(a^2+2a+4)#

#=(a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)#

Explanation:

Use the difference of cubes identity, which can be written:

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

Note that both #a^3# and #8=2^3# are both perfect cubes.

So we find:

#a^3-8#

#=a^3-2^3#

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

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

The remaining quadratic factor has negative discriminant, so no Real zeros and no linear factors with Real coefficients. It is possible to factor it using Complex coefficients:

#a^2+2a+4#

#=a^2+2a+1+3#

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

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

#= (a+1-sqrt(3)i)(a+1+sqrt(3)i)#

So:

#a^8-8 = (a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)#

Another way to express the full factoring is:

#a^3-8 = (a-2)(a-2omega)(a-2omega^2)#

where #omega = -1/2+sqrt(3)/2 i# is the primitive Complex cube root of #1#.

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