Simplifying Rational Expression

[Mi0]

(x−a)(x−b)/(c−b)(c−a)+(x−a)(x−c)/(b−a)(b−c)+(x−b)(x−c)/(a−b)(a−c)
Plz help, ive tried finding common denominator, not help much.This from Gelfand Algebra problem 128

 

[MarkFL]

Hello, and welcome to FMH!

Let's write the expression as:

[MATH]\frac{(x-a)(x-b)}{(a-c)(b-c)}-\frac{(x-a)(x-c)}{(a-b)(b-c)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}[/MATH]
Now, we see the LCD is [MATH](a-b)(a-c)(b-c)[/MATH]
Hence, we get:

[MATH]\frac{(x-a)(x-b)(a-b)-(x-a)(x-c)(a-c)+(x-b)(x-c)(b-c)}{(a-b)(a-c)(b-c)}[/MATH]
What do you get when you expand the numerator, and collect like terms?

 

[Mi0]

You get (a^2b−a^2c−ab^2+ac^2+b^2c−bc^2)/(a-b)(a-c)(b-c)

 

[MarkFL]

Let's just work with the numerator for now...you correctly obtained:

[MATH]a^2b−a^2c−ab^2+ac^2+b^2c−bc^2[/MATH]
Now, what do you get when you expand the denominator?

 

[Mi0]

a^b−a^2c−ab^2+ac^2+b^2c−bc^2

 

[MarkFL]

I think you meant:

[MATH]a^2b−a^2c−ab^2+ac^2+b^2c−bc^2 [/MATH]
And this is correct. So what does the entire rational expression reduce to?

 

[Mi0]

Ah I see it is 1. Thank you very much sir

 

[lookagain]

(x−a)(x−b)/(c−b)(c−a)+(x−a)(x−c)/(b−a)(b−c)+(x−b)(x−c)/(a−b)(a−c)

MiO, for this and similar expressions written in horizontal style, you need added grouping symbols:

(x − a)(x − b)/[(c − b)(c − a)] + (x − a)(x − c)/[(b − a)(b − c)] + (x − b)(x − c)/[(a − b)(a − c)] =

[(x - a)(x - b)(a - b) - (x - a)(x - c)(a - c) + (x - b)(x - c)(b - c)]/[(a - b)(a - c)(b - c)] =

(a^2b − a^2c − ab^2 + ac^2 + b^2c − bc^2)/[(a - b)(a - c)(b - c)] =

(a^2b − a^2c − ab^2 + ac^2 + b^2c − bc^2)/(a^2b − a^2c − ab^2 + ac^2 + b^2c − bc^2) =

1