Algebraic expressions - simplifying

Rationalist
Messages
4
Reaction score
0
Hi all,

I'm retaking the GRE soon and I keep making stupid algebra mistakes in my calculus problems so I'm going through a textbook from 1965 that is helpful but has me stumped on some problems. I hope someone can steer me in the right direction.

1/(a-b)(a-c) + 1/(c-a)(c-b) + 1/(b-a)(b-c)

Is there a way to simplify this? If I start multiplying out everything to get the LCD my final answer will be huge. Thanks in advance
 
Physics news on Phys.org
It shouldn't be too terrible. The obvious thing to note is that (b - a) = -(a - b), for example.
 
Write everything with a denominator of (a-b)(a-c)(b-c) and add the fractions.
 
basic advice: do not fear to write something down, just because you think it will be large and unmanageable. you have to plunge in without fear of what will come out. you have to get your hands dirty.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top