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

thanks..

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

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

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

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

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

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

3. [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?

4. 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!

5. 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)?

6. 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)?

7. $
\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)}$

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