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**Unformatted text preview: **( ) 2 = x 6 . 2 x 6 x 3 1 = 2 u 2 u 1 Now, factor as usual. = 2 u + 1 ( ) u 1 ( ) Substitute back. Replace u with x 3 . = 2 x 3 + 1 ( ) x 3 1 ( ) With practice, the substitution process can be avoided. Either way, we are not done yet! It is true that 2 x 3 + 1 ( ) is prime over the integers; Chapter 2 will help us verify that. However, x 3 1 ( ) is not prime, because we can apply the Difference of Two Cubes formula. 2 x 3 + 1 ( ) x 3 1 ( ) = 2 x 3 + 1 ( ) x 1 ( ) x 2 + x + 1 ( ) This is factored completely over the integers. The Test for Factorability can be used to show that the trinomial factor x 2 + x + 1 ( ) is prime , as expected....

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