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Math Help - need help in exponent problem

  1. #1
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    need help in exponent problem

    if x+1/x = 3^1/2 then x^3 - 1/x^3 = ??

    thanks
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  2. #2
    Member rtblue's Avatar
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    \displaystyle\frac{x+1}{x}=\sqrt{3}

    note that 3^(1/2)=sqrt(3).

    \displaystyle x+1=x\sqrt{3}

    \displaystyle 1=x\sqrt{3}-x

    \displaystyle 1=x(\sqrt{3}-1)

    \displaystyle x=\frac{1}{\sqrt{3}-1}

    Surely, you can take it from here.
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  3. #3
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    Quote Originally Posted by disharmonie View Post
    if x+1/x = 3^1/2 then x^3 - 1/x^3 = ??

    thanks
    Let p = x + \frac{1}{x} = \sqrt{3}.

    Then p^2 = x^2 + \frac{1}{x^2} + 2 \Rightarrow x^2 + \frac{1}{x} = ......

    Let q = x - \frac{1}{x}.

    Then q^2 = ..... = ..... \Rightarrow q = .....

    Now note that x^3 - \frac{1}{x^3} = \left(x^2 + \frac{1}{x^2}\right)q

    The .... are left for you to think about and fill in.
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  4. #4
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    Hello, disharmonie!

    I got a rather silly answer . . . or is it silly?



    \displaystyle \text{If }x+\frac{1}{x} \:=\:\sqrt{3},\,\text{ then }\,x^3 - \frac{1}{x^3} \:=\: ?

    \displaystyle\text{We have: }\;x + \frac{1}{x} \:=\:\sqrt{3}

    \displaystyle \text{Square both sides: }\;\left(x + \frac{1}{x}\right)^2 \:=\:\left(\sqrt{3}\right)^2 \quad\Rightarrow\quad x^2 + 2 + \frac{1}{x^2} \;=\;3<br />

    . . \displaystyle\text{Hence: }\;\boxed{x^2 + \frac{1}{x^2} \;=\;1}


    \displaystyle \text{Subtract 2 from both sides: }\;x^2 - 2 + \frac{1}{x^2} \;=\;-1 \quad\Rightarrow\quad \left(x - \frac{1}{x}\right)^2 \:=\:-1

    . . \displaystyle \text{Hence: }\;\boxed{x - \frac{1}{x} \;=\;\pm i}



    \displaystyle\text{Factor: }\;x^3 - \frac{1}{x^3} \;=\;\left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right)

    . . . . . . . . . . . . . \displaystyle = \;\underbrace{\left(x - \frac{1}{x}\right)}_{\text{This is }\pm i} \left(\underbrace{x^2 + \frac{1}{x^2}}_{\text{This is 1}} + 1\right)

    . . . . . . . . . . . . . =\;\;(\pm i)(2)

    . . . . . . . . . . . . . =\;\;\pm 2i

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