adding algebraic fractions

**princess_21** · January 14th 2009, 12:05 AM

a^3 + b^3 + c^3
---------- ----------- -----------
(a-b)(a-c) (b-a)(b-c) (c-a)(c-b)

thanks..

**red_dog** · January 14th 2009, 01:52 AM

$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$

$=\frac{a^3b-a^3c-ab^3+b^3c+ac^3-bc^3}{(a-b)(a-c)(b-c)}=$

$=\frac{ab(a-b)(a+b)-c(a-b)(a^2+ab+b^2)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$

$=\frac{(a-b)(a^2b+ab^2-a^2c-abc-b^2c+c^3)}{(a-b)(a-c)(b-c)}=$

$=\frac{(a-b)(ab(a-c)+b^2(a-c)-c(a-c)(a+c))}{(a-b)(a-c)(b-c)}=$

$=\frac{(a-b)(a-c)(ab+b^2-ac-c^2)}{(a-b)(a-c)(b-c)}=\frac{(a-b)(a-c)(b-c)(a+b+c)}{(a-b)(a-c)(b-c)}=a+b+c$

**princess_21** · January 14th 2009, 03:08 AM

[quote=red_dog;248305] $\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$

how did you come up with this? on the first step, did you multiply -1 to cancel?

**stapel** · January 14th 2009, 03:26 AM

Originally Posted by princess_21

Originally Posted by red_dog

$\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$

how did you come up with this? on the first step, did you multiply -1 to cancel?

The poster made clever use of an arithmetical fact; no multiplication was necessary.

If you do 5 - 3, you get +2.
If you do 3 - 5, you get -2.

If you "reverse" a subtraction, you get an extra "minus" sign out front. The poster reversed the one subtraction in the denominator (to make the one factor "match" the others) and put the "minus" in front of the entire fraction (rather than leaving it in the denominator or putting it in front only of the numerator; either of these would be "correct", though).

Have fun!

**princess_21** · January 14th 2009, 03:37 AM

on the third expression did you also reversed it? why did you not change the sign? i mean did you reverse only (c-a) not (c-b)?

**stapel** · January 14th 2009, 03:40 AM

Originally Posted by princess_21

on the third expression did you also reversed it? why did you not change the sign? i mean did you reverse only (c-a) not (c-b)?

Was only one of the subtractions reversed in the final result? (Look and count.)

What is the value of (-1)(-1)?

**ADARSH** · January 14th 2009, 03:55 AM

$ \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=I $

Here I is

$ I= \frac{a^3(b-c)}{(a-b)(a-c)(b-c)} -\frac{b^3(a-c)}{(a-b)(a-c)(b-c)} +\frac{c^3(a-b)}{(a-b)(a-c)(b-c)}$

And about R.H.S
$I=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}$

Watch the steps and what was multiplied to it and how (-ve sign) was taken out
Try simplifying and you will get the Right hand side and you will understand it

and ask if still you are in trouble!

**princess_21** · January 14th 2009, 04:45 AM

i got it. thanks.

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