How to factor (x^3 + 1)

- anonymous

How to factor (x^3 + 1)

- schrodinger

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- anonymous

\[(x^{3}+1)\]

- anonymous

mmhm!

- anonymous

no, (x+1)(x^2-x+1)

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## More answers

- Hero

\[(a + b)^3 = (a + b)(a^2 - ab + b^2)\]

- anonymous

\[(x+1)(x^{2}-x+1)\]

- Hero

That's the sum of cubes formula I posted above.

- anonymous

ya basically what they said

- anonymous

ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S

- Hero

No, they are not. One is simply x cubed plus one. The other is \((x + 1) \times (x + 1) \times (x + 1)\)

- anonymous

right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3

- Hero

Because
\((x^3 + 1) = (x^3 + 1^3)\)

- Hero

Remember, \(1 = 1 \times 1 \times 1 = 1^3 \)

- anonymous

= (x + 1) ^3 ?? but wouldnt that make it the same??

- Hero

No, I just explained to you earlier that \((x^3 + 1) \ne (x + 1)^3\)

- Hero

\((x^3 + 1) = (x^3 + 1^3)\)
\((x + 1)^3 = (x + 1)(x + 1)(x + 1)\)

- 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

The only way is if x = 0 or x = 1

- Hero

But in general, they will not be equal

- 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

and the answer is x=17 and x = -1

- anonymous

so how do I get from the sum of cubes to -1

- 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

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

What I say doesn't matter if you don't understand anything.

- Hero

The silence is deafening

- 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

Oh goodie :D

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