How do you factor completely: $49t^6 - 4k^8$ ?

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Answer 1 · 2015-07-13T13:31:22

This is a difference of squares:

$49t^6-4k^8 = (7t^3)^2-(2k^4)^2 = (7t^3-2k^4)(7t^3+2k^4)$

The difference of squares identity is:

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

Use this with $a=7t^3$ and $b=2k^4$ as follows:

$49t^6-4k^8$

$= (7t^3)^2-(2k^4)^2$

$=a^2-b^2$

$= (a-b)(a+b)$

$= (7t^3-2k^4)(7t^3+2k^4)$

How do you factor completely: 49t^6 - 4k^849t^6 - 4k^8?