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

Question

1819 views around the world

You can reuse this answer
Creative Commons License

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

How do you factor 729x^6 - 4096y^6729x^6 - 4096y^6?