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Thread: [SOLVED] Simplifying Radicals

  1. #1
    Member Haven's Avatar
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    [SOLVED] Simplifying Radicals

    Show using Bombelli's Method that $\displaystyle \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 3$

    So I assume the form of the expressions by letting
    $\displaystyle 1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$
    $\displaystyle 2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

    Multiply 1 and 2 and you get: $\displaystyle 1 = a - b^2$
    Cube both sides of 1 and you get: $\displaystyle a^{\frac{3}{2}} + 3ab + 3\sqrt{a}b + b^3 = 18 + \sqrt{325}$
    Which means $\displaystyle a^{\frac{3}{2}} + 3\sqrt{a}{b} = \sqrt{325}$
    and $\displaystyle 3ab + b^3 = 18$

    But I can't get an answer from the equations. Any help would be greatly appreciated
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  2. #2
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    Hello Haven
    Quote Originally Posted by Haven View Post
    Show using Bombelli's Method that $\displaystyle \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 3$

    So I assume the form of the expressions by letting
    $\displaystyle 1. \sqrt[3]{\sqrt{325}+18} = \sqrt{a} + b$
    $\displaystyle 2. \sqrt[3]{\sqrt{325}-18} = \sqrt{a} - b$

    Multiply 1 and 2 and you get: $\displaystyle 1 = a - b^2$
    Cube both sides of 1 and you get: $\displaystyle a^{\frac{3}{2}} + 3ab + 3\sqrt{a}b + b^3 = 18 + \sqrt{325}$
    Which means $\displaystyle a^{\frac{3}{2}} + 3\sqrt{a}{b} = \sqrt{325}$
    and $\displaystyle 3ab + b^3 = 18$

    But I can't get an answer from the equations. Any help would be greatly appreciated
    Bombelli's method is new to me! I should not have thought of using $\displaystyle \sqrt a + b$ and $\displaystyle \sqrt a - b$.

    I agree with your equations, although I think that in order to justify the last two you also need to cube both sides of (2) to get:
    $\displaystyle a^{\frac{3}{2}} - 3ab + 3\sqrt{a}b - b^3 = 18 -\sqrt{325}$
    and then subtract this from your previous equation, before dividing by 2 to get:
    $\displaystyle 3ab + b^3 = 18$
    Then it's a simple matter of eliminating $\displaystyle a$ by substitution:
    $\displaystyle a = b^2+1$

    $\displaystyle \Rightarrow 3(b^2+1)b + b^3 = 188$

    $\displaystyle \Rightarrow 4b^3+3b-18=0$

    $\displaystyle \Rightarrow (2b-3)(2b^2-3b+6)=0$

    $\displaystyle \Rightarrow b = \tfrac32$, since the discriminant of the second factor is negative

    $\displaystyle \Rightarrow \sqrt[3]{\sqrt{325}+18} - \sqrt[3]{\sqrt{325}-18} = 2b$
    $\displaystyle =3$
    Grandad
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  3. #3
    Member Haven's Avatar
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    Thanks.
    Bombelli's method is for simplifying the answers given by Cardano's formulas for cubic equations.
    Cubic function - Wikipedia, the free encyclopedia
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  4. #4
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    Hello Haven
    Quote Originally Posted by Haven View Post
    Thanks.
    Bombelli's method is for simplifying the answers given by Cardano's formulas for cubic equations.
    Cubic function - Wikipedia, the free encyclopedia
    Wow! That all looks pretty fearsome! I think I'll stick to quadratics.

    Grandad
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