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How do you factor #64x^3 - y^6#?

1 Answer

Shwetank Mauria

Jun 15, 2018

#64x^3-y^6=(4x-y^2)(16x^2+4xy^2+y^4)#

Explanation:

As #64x^3-y^6# can be written as #(4x)^3-(y^2)^3#, we can use the identity

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

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

= #(4x-y^2)((4x)^2+4x×y^2+(y^2)^2)#

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

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