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Math Help - adding algebraic fractions

  1. #1
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    Cool adding algebraic fractions

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


    thanks..
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  2. #2
    MHF Contributor red_dog's Avatar
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    \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
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  3. #3
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    [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?
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  4. #4
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    Quote Originally Posted by princess_21 View Post
    Quote Originally Posted by red_dog View Post
    \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!
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  5. #5
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    Question



    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)?
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  6. #6
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    Quote Originally Posted by princess_21 View Post


    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)?
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  7. #7
    Like a stone-audioslave ADARSH's Avatar
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    <br />
\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-a)(b-c)}+\frac{c^3}{(c-a)(c-b)}=I<br />

    Here I is

    <br /> <br />
I= \frac{a^3(b-c)}{(a-b)(a-c)(b-c)} -\frac{b^3(a-c)}{(a-b)(a-c)(b-c)}<br />
+\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!
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  8. #8
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    Thumbs up

    i got it. thanks.
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