How do you factor $x^{6} - 4$ $x^6-4$ ?

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Answer 1 · 2017-05-16T19:31:27

$x^{6} - 4 = (x^{3} + 2) (x^{3} - 2)$ $x^6 -4 = (x^3 +2)(x^3-2)$

Both terms are perfect squares and they are being subtracted.

We have the difference of squares which is factored as :

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

$x^{6} - 4 = (x^{3} + 2) (x^{3} - 2)$ $x^6 -4 = (x^3 +2)(x^3-2)$

Note that $6$ $6$ is not a square, but a base with an even index is a square.

$\sqrt{x^{6}} = x^{3}$ $sqrt(x^6) = x^3$

How do you factor x6−4x^6-4?