Completely factor the difference of cubes
1-x^27

Question

Completely factor the difference of cubes 
1-x^27

anonymous · Answer

$$a^2-b^3=(a-b)(x^2+ab+b^2)$$  in your case $a=1,b=x$

anonymous · Answer

no, in your case $a=1,b=x^3$

loser66 · Answer

* b = x^9

anonymous · Answer

yeah one of those...

anonymous · Answer

so is it suppose to look like this?
1-x^27=(1-x^9)(1+x+x^3)

anonymous · Answer

could anyone help me with my question

anonymous · Answer

During his 2009 MVP season for the St. Louis Cardinals, Albert Pujols hit 93 singles, 45 doubles, 1 triple, and 47 home runs in a total of 568 at bats. The following table arranges these data in terms of the number of bases for a hit, counting 0 bases for the times when he did not get a hit and 4 bases for home runs.

Number of Bases    |     Number of Times
0                            |     382
1                            |     93
2                            |     45
3                            |      1
4                            |     47 
............
it wants to know his typiical bat in 2009 to the nearest thousandth

agent0smith · Answer

I hope for your sake you don't have to completely factor it http://www.wolframalpha.com/input/?i=completely+factor+1-x%5E27 maybe you just have to do it as a difference of two cubes and be done.

agent0smith · Answer

or this http://www4b.wolframalpha.com/Calculate/MSP/MSP84111bha2185cf6b98i100000e821cihh2534968?MSPStoreType=image/gif&s=47&w=440.&h=247.