# Help reducing a fraction to lowest terms...

• February 18th 2013, 12:55 PM
Egoyan
Help reducing a fraction to lowest terms...
Hi guys,

I'm new here, been reading some questions for a while just out of curiosity, learned a thing or two along the way!

I'm doing my homework and I'm stuck on this problem. I need to reduce the following fraction to lowest terms. What do you guys think would be the best way to go about it? I'm getting a bit confused as to where I should start...

https://webwork.mathstat.concordia.c...1e48f9d4e1.png

Thanks a lot guys! I appreciate the help!

Egoyan
• February 18th 2013, 01:04 PM
HallsofIvy
Re: Help reducing a fraction to lowest terms...
Of course, you want to factor those and see if anything in numerator and denominator cancel.

Looking at these, I think you want to use a couple of "well known" products: $x^3- y^3= (x- y)(x^2+ xy+ y^2)$ and $x^2- y^2= (x- y)(x+ y)$.

For example, $(a+ 6)^3- (b- 6)^3= (a+6- (b- 6))((a+6)^2+ (a+ b)(b- 6)+ (b+ 6)^2)$ where I have taken x= a+ 6 and y= b- 6.

And, $(a+ 6)^4- (b- 6)^4= ((a+ 6)^2)^2- ((b- 6)^2)^2= ((a+ 6)^2- (b- 6)^2)((a+ 6)^2+ (b+ 6)^2)$ with $x= (a+ 6)^2$ and $y= (b- 6)^2$. Then, further $(a+ 6)^2- (b- 6)^2= (a+ 6- (b- 6))(a+ 6+ (b- 6))$.
• February 18th 2013, 02:04 PM
Egoyan
Re: Help reducing a fraction to lowest terms...
Ah! I see. Thanks a lot! I'm going to try this way.