Reducing Fractions - (bc-ab) / (a^3 - ac^2 - 2ab^2 + 2b^2c)

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In summary, the conversation is about someone struggling to reduce a fraction and trying to factor out a common term from the denominator. They realize they can factor out (a-c) from (a^2-c^2) to simplify the fraction.
  • #1
lokisapocalypse
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Hey guys,

I'm working on this project for my research and I am having trouble reducing a fraction. The fraction is:

(bc-ab) / (a^3 - ac^2 - 2ab^2 + 2b^2c)

I am almost positive that I can end up with only a (b) on top but I am not sure how. I had an idea:

(b(c-a)) / (a(a^2-c^2) - 2b^2(a - c))

I want to factor out a (a - c) on the bottom but can't because I have (a^2 - c^2).
 
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  • #2
lokisapocalypse said:
Hey guys,

I'm working on this project for my research and I am having trouble reducing a fraction. The fraction is:

(bc-ab) / (a^3 - ac^2 - 2ab^2 + 2b^2c)

I am almost positive that I can end up with only a (b) on top but I am not sure how. I had an idea:

(b(c-a)) / (a(a^2-c^2) - 2b^2(a - c))

I want to factor out a (a - c) on the bottom but can't because I have (a^2 - c^2).

a^2-c^2= (a-c)(a+c), so you can factor out (a-c) from the denominator.
 
  • #3
Thanks, its been a long time since simple stuff like this. You forget about it real quick.
 

FAQ: Reducing Fractions - (bc-ab) / (a^3 - ac^2 - 2ab^2 + 2b^2c)

How can I simplify this fraction?

To simplify this fraction, you can start by factoring out common terms in the numerator and denominator. In this case, you can factor out (b-a) from the numerator and (a^2 - b^2) from the denominator. This will leave you with (b-a) / (a-b)(a^2 - ac - 2b^2 + 2bc).

Can this fraction be reduced any further?

Yes, this fraction can be reduced further. You can factor out (a-b) from the numerator and (a-b) from the denominator, which will leave you with 1 / (a^2 - ac - 2b^2 + 2bc).

What is the simplest form of this fraction?

The simplest form of this fraction is 1 / (a^2 - ac - 2b^2 + 2bc).

Can the terms in the denominator be simplified?

Yes, the terms in the denominator can be simplified further by factoring. You can factor out an a from the first two terms and a b from the last two terms, which will leave you with (a-b)(a-c-2b+2c) in the denominator.

How can I use this fraction in a real-world application?

This fraction can be useful in various situations, such as calculating proportions or ratios, simplifying equations, or solving problems in physics or chemistry. For example, it can be used to calculate the molar ratio of reactants in a chemical reaction or to simplify an equation in a physics problem involving fractions.

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