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