Factoring Help: w^4^m(125r^3+d^3)-81^4^t(125r^3+d^3)

[Norton01]

Getting a different answer every time--

w^4^m(125r^3+d^3)-81^4^t(125r^3+d^3)

What I did:

I broke my 125's down to 5^3's, the same with the 81...ended up with a big mess. I know how to factor cubes, but there are too many going on. And can't figure out if I leave my w^4m's alone, or if I break down the 4. Totally lost. Any help so appreciated.

 

[Loren]

Re: Factoring Help

\(\displaystyle w^4^m(125r^3+d^3)-81^4^t(125r^3+d^3)\)

If the expression were \(\displaystyle w^4^mA-81^4^tA\) would you factor out the A to start?

 

[Aladdin]

Re: Factoring Help

if you want to factorize this equation???
You take (125r^3+d^3)common factor
you get125r^3+d^3)(w^4^m-81^4^t)factorized

 

[Norton01]

Re: Factoring Help

Thanks--
I think what is getting me confused is how far to take the factoring...I mean, do I stop factoring or do I factor 125 and 81? I hate sounding so dense. I think I'm going to leave it as is, and thank for all the help.

 

[Loren]

If the directions are to factor completely, you do so until no further factoring can be done.

 

[Denis]

Re: Factoring Help

Thanks--
I think what is getting me confused is how far to take the factoring...I mean, do I stop factoring or do I factor 125 and 81? I hate sounding so dense. I think I'm going to leave it as is, and thank for all the help.

Only one you can factor now is (w^4^m-81^4^t) ; you realise that's w^(4m) - 81^(4t), right? So factor that.