1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem algebra involving third roots

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let x = third root of [root (108) + 10] - trird root of [root (108) - 10]. Show that x ^ 3 +6 x-20 = 0 from which to infer the value of x (is a small natural)


    3. The attempt at a solution
    may can i have some ideas how to find this? i tryed to find the x but i dont know how with the roots..
     
  2. jcsd
  3. Sep 2, 2012 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Re: Problem algebra

    Hello ParisSpart. Welcome to PF ! (Yes, I see that you have started one other thread, some time ago.)

    What specifically have you tried?

    Where are you stuck?


    To show that [itex]\sqrt[3]{\sqrt{108}+10\ }-\sqrt[3]{\sqrt{108}-10\ }[/itex] is a solution to x3 + 6x - 20 = 0, plug [itex]\sqrt[3]{\sqrt{108}+10\ }-\sqrt[3]{\sqrt{108}-10\ }[/itex] in for x in your equation.

    Of course, you will need to cube [itex]\sqrt[3]{\sqrt{108}+10\ }-\sqrt[3]{\sqrt{108}-10\ }\ .[/itex]

    (a - b)3 = a3 - 3(a2)b + 3a b2 - b3 .
     
  4. Sep 2, 2012 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This is one of those nasty looking problems, but it can be shown that [itex]\sqrt[3]{\sqrt{108}\pm 10\ }=\sqrt{3}\pm 1[/itex].

    Also it may be helpful to express, [itex](a-b)^3[/itex] as
    [itex]a^3-b^3-3ab(a-b)\ .[/itex]​
     
  5. Sep 3, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you know the "cubic formula"? If a and b are any two numbers, then
    [tex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/tex]
    and
    [tex]3ab(a- b)= 3a^2b- 3ab^2[/tex]
    so that [itex](a- b)^3+ 3ab(a- b)= 0[/itex]. That is, if we let x= a- b, m= 3ab, and [itex]n= a^3- b^3[/itex], then [itex]x^3+ mx= n[/itex].

    And, we can do this "the other way around"- knowing m and n, solve for a and b and so solve the (reduced) cubic equation [itex]x^3+ mx= n[/itex]. From m= 3ab, b= m/(3a) and then [itex]n= a^3- m^3/(3^3a^3)[/itex]. Multiplying by [itex]a^3[/itex] gives [itex]na^3= (a^3)^2- (m/3)^3][/itex] which is a quadratic in [itex]a^3[/itex], [itex](a^3)^2- na^3- (m/3)^3= 0[/itex], which can be solved by the quadratic equation:
    [tex]a^3= \frac{n\pm\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{(n/2)^2+ (m3)^3}[/tex]

    Since [itex]a^3- b^= n[/itex],
    [tex]b^3= a^3- n= \frac{n}{2}\mp\sqrt{(n/2)^2+ (m/3)^3}[/tex]

    and x is the difference of cube roots of those.

    The point is that, in this problem, the numbers are given in exactly that form! We can work out that n/2= 10 so n= 20, and that [itex](n/2)^2- (m/3)^3= 100- (m/3)^3= 108[/itex] so that (m/3)^3= -8, m/3= -2, and m= -6.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem algebra involving third roots
Loading...