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Math Help - Solve for x

  1. #1
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    Solve for x

    \sqrt[3]{x^3+98}-x=2

    x should = 3 or -5.

    thanks!
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  2. #2
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    Move -x to the rhs. Cube the rhs and remove the cube root on the lhs. You will probably know how to solve it from there.
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  3. #3
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    Quote Originally Posted by Stuck Man View Post
    Move -x to the rhs. Cube the rhs and remove the cube root on the lhs. You will probably know how to solve it from there.
    Move -x to the rhs.
    \sqrt[3]{x^3+98}=2+x

    Cube the rhs and remove the cube root on the lhs.
    {x^3+98}=2^3+x^3

    do you mean this?

    this cant be right as the x^3 cancel themselves.
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  4. #4
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    No, cube all of the rhs together. You have effectively done that to the lhs.
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  5. #5
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    Quote Originally Posted by Stuck Man View Post
    No, cube all of the rhs together. You have effectively done that to the lhs.
    you mean this?
    <br />
(2+x)^3
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  6. #6
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    Worked it out thanks. common mistake. THanks
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  7. #7
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    Yes. You can't break up the value and then raise the parts to a power.
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