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Math Help - I couldn't solve for a

  1. #1
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    I couldn't solve for a

    When
    a - \frac{1}{a} = 2
    then
    a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} = ?

    The answer is 2

    I tried making a trinomial with a - \frac{1}{a} = 2 and I got a^2 - 2a - 1, but after this I wasn't able to solve for a.

    What should I do to get to the result 2?

    Thanks in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle a^3-2a^2-\frac{2}{a^2}-\frac{1}{a^3}=a^3-\frac{1}{a^3}-2\left(a^2+\frac{1}{a^2}\right)=
    \displaystyle =\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right)-2\left(a^2+\frac{1}{a^2}\right) (1).
    Now, if \displaystyle a-\frac{1}{a}=2 then \displaystyle \left(a-\frac{1}{a}\right)^2=4
    \displaystyle \Rightarrow a^2-2+\frac{1}{a^2}=4\Rightarrow a^2+\frac{1}{a^2}=4+2=6.
    Replacing in (1) we have 2(6+1)-2\cdot 6=2.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Patrick_John View Post
    When
    a - \frac{1}{a} = 2
    then
    a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} = ?

    The answer is 2

    I tried making a trinomial with a - \frac{1}{a} = 2 and I got a^2 - 2a - 1, but after this I wasn't able to solve for a.

    What should I do to get to the result 2?

    Thanks in advance.
    If a - \frac {1}{a} = 2

    Then: a - 2 = \frac {1}{a} ..................(1)
    and : 2 + \frac {1}{a} = a ....................(2)

    Now: a^3 - 2a^2 - \frac {2}{a^2} - \frac {1}{a^3} = a^2 (a - 2) - \frac {1}{a^2} \left( 2 + \frac {1}{a} \right) .............I factored out a^2 from the first two terms and - \frac {1}{a^2} from the last two terms.

    Now we can replace what's in brackets above with equations (1) and (2) respectively

    \Rightarrow a^3 - 2a^2 - \frac {2}{a^2} - \frac {1}{a^3} = a^2 \left( \frac {1}{a} \right) - \frac {1}{a^2} (a)
    ................................ = a - \frac {1}{a}

    ................................ = 2


    EDIT: Beaten by the great red_dog. Nice Solution, red_dog!
    Last edited by Jhevon; June 28th 2007 at 09:59 PM.
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  4. #4
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    Hello, Patrick_John!

    When a - \frac{1}{a} \:= \:2, then: . a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} \:= \:?

    The answer is 2

    Square the equation: . \left(a - \frac{1}{a}\right)^2 \:=\:2^2

    We have: . a^2 - 2 + \frac{1}{a^2} \:=\:4\quad\Rightarrow\quad a^2 + \frac{1}{a^2} \:=\:6

    Multiply by \text{-}2\!:\;\;\text{-}2a^2 - \frac{2}{a^2} \:=\:\text{-}12 .[1]


    Cube the equation: . \left(a - \frac{1}{a}\right)^3 \:=\:2^3

    We have: . a^3 - 3a + \frac{3}{a} - \frac{1}{a^3} \:=\:8\quad\Rightarrow\quad a^3 - 3\underbrace{\left(a - \frac{1}{a}\right)}_{\text{this is 2}} - \frac{1}{a^3} \:=\:8<br />

    . . . .Hence we have: . a^3 - \frac{1}{a^3} \;=\;14

    . . . .Add [1]:. . . . . \text{-}2a^2 - \frac{2}{a^2} \;=\;\text{-}12


    Therefore: . a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} \;=\;2

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  5. #5
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    Wow! Thanks a lot guys!

    Can anybody recommend me a site or something where I can read more about this topic?
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