In my endless assignment of factoring questions, I've come across another that I can't quite get:

3(x+2w)^3 - 3p^3r^3

Thanks for any help :)

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- Jul 8th 2007, 05:46 PMstarsweptFactoring question
In my endless assignment of factoring questions, I've come across another that I can't quite get:

3(x+2w)^3 - 3p^3r^3

Thanks for any help :) - Jul 8th 2007, 05:59 PMJhevon
$\displaystyle 3(x + 2w)^3 - 3p^3 r^3$ .........factor out the 3

$\displaystyle \Rightarrow 3 \left( (x + 2w)^3 - (pr)^3 \right)$

Now the inside of the brackets is the difference of two cubes, which you learned a formula for in class recently (i think), so you can take it from here

(if not say so) - Jul 8th 2007, 06:53 PMstarswept
Thanks again Jhevon :) I factored out the three, and used the formula I know, which is (a^3-b^3) = (a-b)(a^2+ab+b^2).

I ended up with 3(x+2w-pr)(x^2-prx-2prw+p^2r^2). However, the answer should, according to the text, be 3(x+2x-pr)(x^2 + 4xw + 4w^2 + prx + prw + p^2r^2).

This textbook is notorious for having mistakes in the answers, but it can't be that off, so I must have made a mistake with the formula somewhere, but I can't spot it. Can anyone point out my error? Thanks in advance! - Jul 8th 2007, 07:02 PMtopsquark
There is a slight problem in the book's answer, but I suspect it's typos on your part. (See the red x above and the coefficient on the wpr term.) I'd say you expanded wrong somewhere, but I can't figure out where.

$\displaystyle 3(x+2w)^3 - 3p^3r^3$

$\displaystyle = 3[(x + 2w)^3 - p^3r^3] $

$\displaystyle = 3[(x + 2w) - (pr)][(x + 2w)^2 + (x + 2w)(pr) + (pr)^2] $

$\displaystyle = 3(x + 2w - pr)(x^2 + 4xw + 4w^2 + xpr + 2wpr + p^2r^2)$

-Dan - Jul 8th 2007, 07:03 PMJhevon
- Jul 8th 2007, 07:04 PMJhevon
- Jul 8th 2007, 07:21 PMstarswept
I see my mistake. I conveniently squared only the x of (x+2w), and forgot the 2w existed. Stupid errors like that are why taking accelerated calculus after no math for six months is dangerous :rolleyes: And as sad as it is, Dan, those errors in the book's equation were not typos, but from the book straight-on. This book has so many errors that having the answers can be more of a hindrance than a help. It makes getting the right answer a fun game of telephone tag for people in my math class ;)

Anyways, thanks again for the help, both of you! - Jul 8th 2007, 07:22 PMtopsquark
- Jul 8th 2007, 07:25 PMJhevon
- Jul 8th 2007, 07:32 PMtopsquark