One obvious root of this is x = 1, so (x-1) is a factor. Use long division of (x3 -1)/(x-1) and the quotient (x2 + x + 1) will be another factor. So (x3 -1) = (x-1)(x2 + x + 1). Remember
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Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 + ab+b2) a 3 - b 3 = ( a - b) ( a 2 + a b + b 2) where a = x a = x and b = 3 b = 3. (x−3)(x2
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