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Brittni0605 Group TitleBest ResponseYou've already chosen the best response.0
\[(x^{3}+1)\]
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cutiepo0 Group TitleBest ResponseYou've already chosen the best response.1
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)(x17)
 one year ago

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

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

Hero Group TitleBest ResponseYou've already chosen the best response.1
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 \((x^3 + 1)(x  17) = (x + 1)(x^2  x + 1)(x  17) = 0\)
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

Hero Group TitleBest ResponseYou've already chosen the best response.1
The silence is deafening
 one year ago

Cutiepo0 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago
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