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Thread: prove the following

  1. #1
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    prove the following

    IF: X + 1/X = 48
    and X + 1/X = 7

    then prove that X + 1/X = 329
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  2. #2
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    Re: prove the following

    Do you know how to factorise the sum of two cubes?
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  3. #3
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    Re: prove the following

    Quote Originally Posted by neelsvanas View Post
    IF: X + 1/X = 48
    and X + 1/X = 7

    then prove that X + 1/X = 329
    $\displaystyle \begin{align*}{x^3} + \frac{1}{{{x^3}}} &= \left( {x + \frac{1}{x}} \right)\left( {{x^2} - x\left( {\frac{1}{x}} \right) + \frac{1}{{{x^2}}}} \right)\\& = 7\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)\\& = 7\left( {48 - 1} \right)\\& = ~? \end{align*}$
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  4. #4
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    Re: prove the following

    the given system of equations is inconsistent

    $\displaystyle 49=7^2=\left(x+\frac{1}{x}\right)^2=2+\frac{1}{x^2 }+x^2=50$
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  5. #5
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    Re: prove the following

    Quote Originally Posted by Idea View Post
    the given system of equations is inconsistent

    $\displaystyle 49=7^2=\left(x+\frac{1}{x}\right)^2=2+\frac{1}{x^2 }+x^2=50$
    neelsvanas,

    you see what is in the quote box? Idea pointed out that the solutions for your first equation are not that for your
    second equation in post #1.

    Your problem is wrong (with a couple of numbers). Here is a correct restatement:

    If $\displaystyle \ x^2 + \dfrac{1}{x^2} \ = \ $ 47

    and $\displaystyle \ x + \dfrac{1}{x} \ = \ $ 7,

    then prove that $\displaystyle \ x^3 + \dfrac{1}{x^3} \ = \ $ 322.


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    A more challenging problem would have been:

    If $\displaystyle \ x + \dfrac{1}{x} \ = \ $ 7,

    then determine the value of $\displaystyle \ \ x^3 + \dfrac{1}{x^3} $.
    Last edited by greg1313; Apr 3rd 2019 at 07:34 AM.
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  6. #6
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    Re: prove the following

    Thanks from topsquark
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