anonymous
  • anonymous
How to factor (x^3 + 1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[(x^{3}+1)\]
anonymous
  • anonymous
mmhm!
anonymous
  • anonymous
no, (x+1)(x^2-x+1)

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More answers

Hero
  • Hero
\[(a + b)^3 = (a + b)(a^2 - ab + b^2)\]
anonymous
  • anonymous
\[(x+1)(x^{2}-x+1)\]
Hero
  • Hero
That's the sum of cubes formula I posted above.
anonymous
  • anonymous
ya basically what they said
anonymous
  • anonymous
ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S
Hero
  • Hero
No, they are not. One is simply x cubed plus one. The other is \((x + 1) \times (x + 1) \times (x + 1)\)
anonymous
  • anonymous
right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3
Hero
  • Hero
Because \((x^3 + 1) = (x^3 + 1^3)\)
Hero
  • Hero
Remember, \(1 = 1 \times 1 \times 1 = 1^3 \)
anonymous
  • anonymous
= (x + 1) ^3 ?? but wouldnt that make it the same??
Hero
  • Hero
No, I just explained to you earlier that \((x^3 + 1) \ne (x + 1)^3\)
Hero
  • Hero
\((x^3 + 1) = (x^3 + 1^3)\) \((x + 1)^3 = (x + 1)(x + 1)(x + 1)\)
anonymous
  • anonymous
ok, so (a^3 + b^3) doesn't equal (a+b)^3, ever right? Sorry, just trying to wrap my head around it
Hero
  • Hero
The only way is if x = 0 or x = 1
Hero
  • Hero
But in general, they will not be equal
anonymous
  • anonymous
ok, 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)(x-17)
anonymous
  • anonymous
and the answer is x=17 and x = -1
anonymous
  • anonymous
so how do I get from the sum of cubes to -1
Hero
  • Hero
Basically, 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
  • Hero
If 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
  • Hero
What I say doesn't matter if you don't understand anything.
Hero
  • Hero
The silence is deafening
anonymous
  • anonymous
Oh, 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
Hero
  • Hero
Oh goodie :D

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