How do you factor $x^6 - 64$ ?

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Answer 1 · 2015-04-15T16:34:23

(x+2)(x-2)( $x^4$ +4 $x^2$ +16)

Rewrte the given terms as perfect cubes as follows and use the algebraic identity for the factorisation of difference of two cubes.

$x^6$ -64 = $(x^2)^3$ - $(4)^3$

=( $x^2$ -4) ( $x^4$ +4 $x^2$ +16)

=(x+2)(x-2)( $x^4$ +4 $x^2$ +16)

Answer 2 · 2015-04-15T18:14:14

This is a bit of a tricky question.

If you write $x^6-64 = (x^2)^3 - 4^3$ you can factor it into:

$(x+2)(x-2)(x^4 +4x^2+16)$

If you write:
$x^6-64 = (x^3)^2 - 8^2$ , then you factor as

$(x^3+8)(x^3-8)$ which can be further factored as:

$(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$

The second answer is factored into irreducibles (over $RR$ ).

How do you factor x^6 - 64x^6 - 64?