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bandicoot12
 one year ago
Best ResponseYou've already chosen the best response.0no, (x+1)(x^2x+1)

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

Brittni0605
 one year ago
Best ResponseYou've already chosen the best response.0\[(x+1)(x^{2}x+1)\]

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

bandicoot12
 one year ago
Best ResponseYou've already chosen the best response.0ya basically what they said

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S

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

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3

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

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

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1= (x + 1) ^3 ?? but wouldnt that make it the same??

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

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1ok, so (a^3 + b^3) doesn't equal (a+b)^3, ever right? Sorry, just trying to wrap my head around it

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

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

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1ok, 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)

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1and the answer is x=17 and x = 1

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1so how do I get from the sum of cubes to 1

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

Cutiepo0
 one year ago
Best ResponseYou've already chosen the best response.1Oh, 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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