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Thread: I couldn't solve for a

  1. #1
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    I couldn't solve for a

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

    The answer is $\displaystyle 2$

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

    What should I do to get to the result $\displaystyle 2$?

    Thanks in advance.
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \displaystyle a^3-2a^2-\frac{2}{a^2}-\frac{1}{a^3}=a^3-\frac{1}{a^3}-2\left(a^2+\frac{1}{a^2}\right)=$
    $\displaystyle \displaystyle =\left(a-\frac{1}{a}\right)\left(a^2+1+\frac{1}{a^2}\right)-2\left(a^2+\frac{1}{a^2}\right)$ (1).
    Now, if $\displaystyle \displaystyle a-\frac{1}{a}=2$ then $\displaystyle \displaystyle \left(a-\frac{1}{a}\right)^2=4$
    $\displaystyle \displaystyle \Rightarrow a^2-2+\frac{1}{a^2}=4\Rightarrow a^2+\frac{1}{a^2}=4+2=6$.
    Replacing in (1) we have $\displaystyle 2(6+1)-2\cdot 6=2$.
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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
    $\displaystyle a - \frac{1}{a} = 2$
    then
    $\displaystyle a^3 - 2a^2 - \frac{2}{a^2} - \frac{1}{a^3} = ? $

    The answer is $\displaystyle 2$

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

    What should I do to get to the result $\displaystyle 2$?

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

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

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

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

    $\displaystyle \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) $
    ................................$\displaystyle = a - \frac {1}{a}$

    ................................$\displaystyle = 2$


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

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

    The answer is $\displaystyle 2$

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

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

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


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

    We have: .$\displaystyle 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
    $

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

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


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

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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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