$\displaystyle

\frac {x^2 - 27}{x - 3}

$

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- Apr 7th 2009, 09:15 PMTenskypooSimplifying: difference of two cubes
$\displaystyle

\frac {x^2 - 27}{x - 3}

$ - Apr 7th 2009, 09:16 PMJhevon
- Apr 7th 2009, 09:22 PMTenskypoo
yup. I've gotten to where it is

(x+27)(x^2 - (x times 27))+ 27^2

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x-3

- Apr 7th 2009, 09:25 PMJhevon
- Apr 7th 2009, 09:30 PMTenskypoo
ah, because 3 x 3 x 3= 27

So if it was, x^4 - 16 it would be x^4 - 2^4 just as an example to see if i understand. - Apr 7th 2009, 09:34 PMJhevon
- Apr 7th 2009, 09:35 PMTenskypoo
- Apr 7th 2009, 09:36 PMJhevon
- Apr 7th 2009, 09:38 PMTenskypoo
i'd put that over x-3, but cancel out the x-3 since it is in both the numer. and denom. therefore leaving me with

$\displaystyle x^2 + 3x + 9$

i may have done the part wrong, or it isn't simplified enough - Apr 7th 2009, 09:39 PMJhevon
- Apr 8th 2009, 05:10 AMstapel
Make sure you learn and

*memorize*the formula for factoring a**difference of cubes**before the next test! (Wink)