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Thread: Surds?

  1. #1
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    Surds?

    Please help me with the following.....

    Simplify

    ((108^1/2) + 10)^1/3 - ((108^1/2) - 10)^1/3

    I gave it many tries but still got no headstart.
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  2. #2
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    Quote Originally Posted by madaboutmath View Post
    Please help me with the following.....

    Simplify

    ((108^1/2) + 10)^1/3 - ((108^1/2) - 10)^1/3

    I gave it many tries but still got no headstart.
    Very you solutions cubics algebraically?

    Because let $\displaystyle x$ be this number. Cube both sides by using the fact $\displaystyle (a-b)^3 = a^3 - 3ab(a-b) - b^3$.
    This gives us,
    $\displaystyle x^3 = 20 - 3\sqrt[3]{8}x$
    Thus,
    $\displaystyle x^3 + 6x = 20$.
    It is 'easy' to see that $\displaystyle x=2$ is a solution.
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  3. #3
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    I couldn't get the -3x(8)^1/3. please help ThePerfectHacker.

    Thanks.
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  4. #4
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    Quote Originally Posted by madaboutmath View Post
    I couldn't get the -3x(8)^1/3. please help ThePerfectHacker.

    Thanks.
    Let $\displaystyle a$ be the first radical and $\displaystyle b$ be the second radical. So you want to find the value of $\displaystyle a-b$. Let $\displaystyle x=a-b$ when you cube both sides you get, $\displaystyle x^3 = a^3 - 3ab(a-b) - b^2$. Thus, $\displaystyle x^3 = a^3 - b^3 - 3abx$. Now $\displaystyle a^3 - b^3 = 20$, that is easy to see because when you cube them the cube root goes away and the square roots cancel. And $\displaystyle ab$ is the difference of two square because $\displaystyle ab = \sqrt[3]{\sqrt{108}+10}\cdot \sqrt[3]{\sqrt{108}-10} = \sqrt[3]{108-100} = 2$.
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  5. #5
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    Got it already! Thanks so much!

    MAM
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