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Radicals with order higher than 2

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data
    order 3 radical from (9+4radical5) + order 3 radical from (9-4radical5)

    *note: 9+4radical5 is all under the big order 3 radical. (so is the other one)

    2. Relevant equations
    in case you don't understand what i wrote (sorry, my language is not english and math linguistics differ in my language) click the attachement to see a picture of the equation


    3. The attempt at a solution
    I think i might find the solution by writing 9+4radical5 as a cube of something so as to cancel the radical but what if i have order 6 radical? (because i have this type of exercises as the next one. So clearly writing as a square, cube etc won't work everytime. Thank you!
     

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    Last edited: Jun 5, 2012
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  3. Jun 5, 2012 #2

    SammyS

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    Hallo Eurovision. Welcome to PF !

    I take it that you need to in some way simplify [itex]\displaystyle \sqrt[3]{9+4\sqrt{5}\ }+\sqrt[3]{9-4\sqrt{5}\ }\,.[/itex] Is that correct?
     
  4. Jun 5, 2012 #3
    Thank you.

    Yes, i must find what it equals to (I know already is 3 from the answers) but i don't know how to solve it.
     
  5. Jun 5, 2012 #4
  6. Jun 5, 2012 #5

    HallsofIvy

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    I would do this is in a rather different way. Cardano's formula for the "reduced" cubic, [itex]x^3+ px+ q= 0[/itex], gives a result of the form
    [tex]x= \sqrt[3]{-\frac{q}{2}+ \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+ \sqrt[3]{-\frac{q}{2}- \sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}}[/tex]

    The number in this problem, being a sum of cube roots, is of that form with -q/2= 9 and [itex]\sqrt{q^2/4+p^3/27}= 4\sqrt{5}[/itex] so that [itex]q^2/4+ p^3/27= 80[/itex]. Solving for p and q gives you a cubic equation that has x= 3 as its only real root.
     
  7. Jun 5, 2012 #6
    Interesting method. *like* :cool:
     
  8. Jun 5, 2012 #7
    Here's how I would do it: cube the entire thing.

    Something helpful: If you have [itex]S = A + B[/itex], then [itex]S^3 = A^3 + B^3 + 3AB(A + B)[/itex]. And what is [itex]A + B[/itex]? :)
     
  9. Jun 5, 2012 #8

    SammyS

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    Notice that [itex]\displaystyle \sqrt[3]{9+4\sqrt{5}\ }+\sqrt[3]{9-4\sqrt{5}\ }\,[/itex]


    can be written as [itex]\displaystyle \sqrt[3]{\sqrt{81}+\sqrt{80}\ }+\sqrt[3]{\sqrt{81}-\sqrt{80}}\ .[/itex]
     
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