a^3 + b^3 + c^3
---------- ----------- -----------
(a-b)(a-c) (b-a)(b-c) (c-a)(c-b)
thanks..
$\displaystyle \frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$
$\displaystyle =\frac{a^3b-a^3c-ab^3+b^3c+ac^3-bc^3}{(a-b)(a-c)(b-c)}=$
$\displaystyle =\frac{ab(a-b)(a+b)-c(a-b)(a^2+ab+b^2)+c^3(a-b)}{(a-b)(a-c)(b-c)}=$
$\displaystyle =\frac{(a-b)(a^2b+ab^2-a^2c-abc-b^2c+c^3)}{(a-b)(a-c)(b-c)}=$
$\displaystyle =\frac{(a-b)(ab(a-c)+b^2(a-c)-c(a-c)(a+c))}{(a-b)(a-c)(b-c)}=$
$\displaystyle =\frac{(a-b)(a-c)(ab+b^2-ac-c^2)}{(a-b)(a-c)(b-c)}=\frac{(a-b)(a-c)(b-c)(a+b+c)}{(a-b)(a-c)(b-c)}=a+b+c$
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!
$\displaystyle
\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
$\displaystyle
I= \frac{a^3(b-c)}{(a-b)(a-c)(b-c)} -\frac{b^3(a-c)}{(a-b)(a-c)(b-c)}
+\frac{c^3(a-b)}{(a-b)(a-c)(b-c)}$
And about R.H.S
$\displaystyle I=\frac{a^3(b-c)-b^3(a-c)+c^3(a-b)}{(a-b)(a-c)(b-c)}$
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!