How do you factor #x^6 - 64#?

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Answer 1 · 2015-04-15T16:34:23

(x+2)(x-2)(#x^4# +4#x^2#+16)

Rewrte the given terms as perfect cubes as follows and use the algebraic identity for the factorisation of difference of two cubes.

#x^6# -64 = #(x^2)^3# - #(4)^3#

=(#x^2# -4) (#x^4# +4#x^2#+16)

=(x+2)(x-2)(#x^4# +4#x^2#+16)

Answer 2 · 2015-04-15T18:14:14

This is a bit of a tricky question.

If you write #x^6-64 = (x^2)^3 - 4^3# you can factor it into:

#(x+2)(x-2)(x^4 +4x^2+16)#

If you write:
#x^6-64 = (x^3)^2 - 8^2# , then you factor as

#(x^3+8)(x^3-8)# which can be further factored as:

#(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)#

The second answer is factored into irreducibles (over #RR#).