Factor #x^6 - 64# over Real numbers?

1 Answer
George C.
Feb 10, 2018

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

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Hence we find:

#x^6-64 = (x^3)^2-8^2#

#color(white)(x^6-64) = (x^3-8)(x^3+8)#

#color(white)(x^6-64) = (x^3-2^3)(x^3+2^3)#

#color(white)(x^6-64) = (x-2)(x^2+2x+4)(x+2)(x^2-2x+4)#

Notes

It is interesting to see what happens if you use the difference of cubes identity first:

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

#color(white)(x^6-64) = (x^2-4)((x^2)^2+4x^2+4^2)#

#color(white)(x^6-64) = (x^2-2^2)((x^2+4)^2-(2x)^2)#

#color(white)(x^6-64) = (x-2)(x+2)((x^2+4)-2x)((x^2+4)+2x)#

#color(white)(x^6-64) = (x-2)(x+2)(x^2-2x+4)(x^2+2x+4)#

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