What are the factors of #x^3y^6 – 64#?

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Answer 1 · 2014-10-29T00:58:45

#x^3y^6 - 64# is the difference of two cubes and can be factored in the following pattern.

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

#a^3# factors to a
#b^3# factors to b

The pattern of the signs follows the acronym SOAP
S = same sign as the cubes
O = opposite sins of the cubes
AP = always positive

#x^3y^3# factors to xy
#64# factors to 4

#x^3y^3 - 64 = (xy - 4)(x^2y^2 + 4xy + 16)#

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