Math problem?

Question

I have to simplify this problem..
http://i3.photobucket.com/albums/y90/lazydaisy2828/precalcprob.jpg
I got it sorta simplified as x with a really big fractional exponent... I dont think I did it right..
See what you get, and if you get something, please explain.. :) thanks

Answer 1

It is x to this exponent

a^2(a+b) + 2abc + b^2(c+a) + c^2(a+b+c) over
(c-a)^2 * (b-c)^2 * (a-b)^2

I multiplied out the denomenator but nothing cancels.

Answer 2

Deal with the exponents only:

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

The common denominator is (b-c)*(a-b)*(c-a)

The numerator is (a+b)*(a-b) + (c+a)*(c-a) + (b+c)*(b-c);
remember that (a+b)*(a-b) = a^2-b^2; similarly for (c+a)*(c-a) and (b+c)*(b-c); You will get numerator terms of a^2,b^2, and c^2, some (maybe all) of which will cancel out.

Answer 3

What you need to do is to multiply each pair of exponents, then you'll add the exponents together.

x^{(a + b)/[(c - a)(b - c)] + (c + a)/[(b - c)(a - b)] + (b + c)/[(a - b)(c - a)]}

The common denominator: (a - b)(b - c)(c - a)

So all you need to do now is to convert each fraction into the common denominator and add them.

Answer 4

sorry...im not in pre-caculus..im only in algebra...1

Answer 5

No idea!