How do you factor #x^6-4#?

1 Answer
EZ as pi
May 16, 2017

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

Explanation:

Both terms are perfect squares and they are being subtracted.

We have the difference of squares which is factored as :

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

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

Note that #6# is not a square, but a base with an even index is a square.

#sqrt(x^6) = x^3#

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