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Math Help - Algebraic expressions

  1. #1
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    Algebraic expressions

    Hello everyone.

    I got the following problem:

    If a *b != 0 and a != b, then the following expression:

    (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}

    equals:
    a) a^2+ab+b^2
    b) a-b
    c) a+b
    d) \frac{1}{a}-\frac{1}{b}
    e) \frac{1}{a}+\frac{1}{b}

    So, I started like this:

    (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}

    (\frac{a^2-2ab+b^2}{ab}+3)*(\frac{a^2}{ab}-\frac{b^2}{ab})*\frac{ab}{a^3-b^3}

    (\frac{a^2-2ab+b^2}{ab}+3)*\frac{a^2-b^2}{a^3-b^3}

    and that's where I'm stuck
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  2. #2
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    Quote Originally Posted by ikislav View Post
    Hello everyone.

    I got the following problem:

    If a *b != 0 and a != b, then the following expression:

    (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}

    equals:
    a) a^2+ab+b^2
    b) a-b
    c) a+b
    d) \frac{1}{a}-\frac{1}{b}
    e) \frac{1}{a}+\frac{1}{b}

    So, I started like this:

    (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}

    (\frac{a^2-2ab+b^2}{ab}+3)*(\frac{a^2}{ab}-\frac{b^2}{ab})*\frac{ab}{a^3-b^3}

    (\frac{a^2-2ab+b^2}{ab}+3)*\frac{a^2-b^2}{a^3-b^3}

    and that's where I'm stuck

    I assume the following is the correct problem statement ...

    If a \cdot b \ne = 0 and a \ne b, then the following expression:

    (\frac{(a-b)^2}{ab}+3) \cdot (\frac{a}{b}-\frac{b}{a}) \div \frac{a^3-b^3}{ab}

    equals:
    a) a^2+ab+b^2
    b) a-b
    c) a+b
    d) \frac{1}{a}-\frac{1}{b}
    e) \frac{1}{a}+\frac{1}{b}

    ...

    \left[\frac{a^2 - 2ab + b^2}{ab}+3\right] \cdot \frac{a^2-b^2}{a^3-b^3}

    \left[\frac{a^2 - 2ab + b^2}{ab}+\frac{3ab}{ab}\right] \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}

    \frac{a^2 +ab + b^2}{ab} \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}

    can you finish?
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  3. #3
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    Quote Originally Posted by ikislav View Post
    If a *b != 0 and a != b, then the following expression: (\frac{(a-b)^2}{ab}+3)*(\frac{a}{b}-\frac{b}{a}):\frac{a^3-b^3}{ab}
    Note two things: \left( {\dfrac{{a^2  - 2ab + b^2 }}{{ab}} + 3} \right) = \left( {\dfrac{{a^2  + ab + b^2 }}{{ab}}} \right)

    a^3-b^3=(a-b)(a^2+ab+b^2)~.
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  4. #4
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    Quote Originally Posted by skeeter View Post
    I assume the following is the correct problem statement ...

    If a \cdot b \ne = 0 and a \ne b, then the following expression:

    (\frac{(a-b)^2}{ab}+3) \cdot (\frac{a}{b}-\frac{b}{a}) \div \frac{a^3-b^3}{ab}

    equals:
    a) a^2+ab+b^2
    b) a-b
    c) a+b
    d) \frac{1}{a}-\frac{1}{b}
    e) \frac{1}{a}+\frac{1}{b}

    ...

    \left[\frac{a^2 - 2ab + b^2}{ab}+3\right] \cdot \frac{a^2-b^2}{a^3-b^3}

    \left[\frac{a^2 - 2ab + b^2}{ab}+\frac{3ab}{ab}\right] \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}

    \frac{a^2 +ab + b^2}{ab} \cdot \frac{(a-b)(a+b)}{(a-b)(a^2+ab+b^2)}

    can you finish?
    Ok, so than that's:

    \frac{a+b}{ab}

    Is that right? But it is not among the given choice of answers.
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  5. #5
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    Quote Originally Posted by ikislav View Post
    Ok, so than that's:

    \frac{a+b}{ab}

    Is that right? But it is not among the given choice of answers.
    yes, it is ...
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  6. #6
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    Quote Originally Posted by skeeter View Post
    yes, it is ...
    Ok, if I multiply with ab, I get a^2b+ab^2. I am at a loss here.
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  7. #7
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    \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} = ... one more step
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  8. #8
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    Quote Originally Posted by skeeter View Post
    \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} = ... one more step
    I got it:

    \dfrac{a+b}{ab} = \dfrac{a}{ab} + \dfrac{b}{ab} = \dfrac{1}{b} + \dfrac{1}{a}

    Right answer is the one under "e"
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