How do you factor $8 x^{3} - 8$ $8x^3-8$ ?

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Answer 1 · 2015-05-25T22:10:44

$8 x^{3} - 8 = 8 (x^{3} - 1) = 8 (x^{3} - 1^{3})$ $8x^3-8 = 8(x^3-1) = 8(x^3-1^3)$

$= 8 (x - 1) (x^{2} + x + 1)$ $= 8(x-1)(x^2+x+1)$

Using the difference of cubes identity
$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

The quadratic factor can only be split into linear factors with complex coefficients.

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