How do you factor $(x^6)+(7x^3)-8$ ?

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Answer 1 · 2017-02-03T22:49:31

$x^6+7x^3-8 = (x+2)(x^2-2x+4)(x-1)(x^2+x+1)$

The sum of cubes identity can be written:

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

The difference of cubes identity can be written:

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

So we find:

$x^6+7x^3-8 = (x^3+8)(x^3-1)$

$color(white)(x^6+7x^3-8) = (x^3+2^3)(x^3-1^3)$

$color(white)(x^6+7x^3-8) = (x+2)(x^2-2x+4)(x-1)(x^2+x+1)$

How do you factor (x^6)+(7x^3)-8(x^6)+(7x^3)-8?