How do you factor #729x^6 - 4096y^6#?

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Answer 1 · 2016-05-09T02:05:04

#(3x-4y)(3x+4y)(9x^2+12xy+16y^2)#

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

Here, #729x^6-4096y^6#

#=(3x)^6-(4y)^6#

#=(a^3-b^3)#, where #a=(3x)^2 and b=(4y)^2#

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

Now, use #(a^2-b^2)=(a-b)(a+b)#

So, #729x^6-4096y^6=(3x-4y)(3x+4y)(9x^2+12xy+16y^2)#