My Math Forum Factoring problem: x^3 -1

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 January 5th, 2012, 01:09 AM #1 Newbie   Joined: Nov 2011 Posts: 21 Thanks: 0 Factoring problem: x^3 -1 I know from my book that x^3 - 1 factors into: (x-1)(x^2+x+1) but I don't know how they got there. Another example is: x^3 - 8 = x^2 + 2x + 4 . What's the logic here? Thank you.
 January 5th, 2012, 02:00 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,279 Thanks: 1023 Re: Factoring problem: x^3 -1
 January 5th, 2012, 02:22 AM #3 Newbie   Joined: Nov 2011 Posts: 21 Thanks: 0 Re: Factoring problem: x^3 -1 thanks
 January 5th, 2012, 04:30 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,929 Thanks: 1124 Math Focus: Elementary mathematics and beyond Re: Factoring problem: x^3 -1 Difference of cubes: $(a\,-\,b)^3\,=\,a^3\,-\,3a^2b\,+\,3ab^2\,-\,b^3$ \begin{align*}a^3\,-\,b^3\,&=\,(a\,-\,b)^3\,+\,3a^2b\,-\,3ab^2 \\ &=\,(a\,-\,b)^3\,+\,3ab(a\,-\,b) \\ &=\,(a\,-\,b)((a\,-\,b)^2\,+\,3ab) \\ &=\,(a\,-\,b)(a^2\,+\,ab\,+\,b^2)\end{align*} The formula for the sum of cubes can be derived in a similar fashion.

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