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anonymous
 3 years ago
How to factor (x^3 + 1)
anonymous
 3 years ago
How to factor (x^3 + 1)

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Hero
 3 years ago
Best ResponseYou've already chosen the best response.1\[(a + b)^3 = (a + b)(a^2  ab + b^2)\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1That's the sum of cubes formula I posted above.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya basically what they said

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1No, they are not. One is simply x cubed plus one. The other is \((x + 1) \times (x + 1) \times (x + 1)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1Because \((x^3 + 1) = (x^3 + 1^3)\)

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1Remember, \(1 = 1 \times 1 \times 1 = 1^3 \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0= (x + 1) ^3 ?? but wouldnt that make it the same??

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1No, I just explained to you earlier that \((x^3 + 1) \ne (x + 1)^3\)

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1\((x^3 + 1) = (x^3 + 1^3)\) \((x + 1)^3 = (x + 1)(x + 1)(x + 1)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, so (a^3 + b^3) doesn't equal (a+b)^3, ever right? Sorry, just trying to wrap my head around it

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1The only way is if x = 0 or x = 1

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1But in general, they will not be equal

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, so now though with the sum of cubes formula, the whole point was to find the zeroes of the equation f(x)= (x^3 + 1)(x17)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the answer is x=17 and x = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so how do I get from the sum of cubes to 1

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1Basically, the question is asking: What x values would make \((x^3 + 1)(x  17) = 0\) You know that anything times zero equals zero. So when you expand \((x^3 + 1)\) you get \((x + 1)(x^2  x + 1)\) Thus \((x^3 + 1)(x  17) = (x + 1)(x^2  x + 1)(x  17) = 0\)

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1If you look closely you'll realize that if (x + 1) = 0 or if (x  17) = 0, then the product of the whole thing will be zero. There's only one way to make (x + 1) = 0 There's only one way to make (x  17) = 0.

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1What I say doesn't matter if you don't understand anything.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, I got it, thanks a lot for your help with this :) b/c (x2−x+1) doesn't have any roots, so it's just the other two roots
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