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Problem algebra involving third roots

  • Thread starter ParisSpart
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Homework Statement


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)


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..
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Let x = third root of [root (108) + 10] - third 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)

The Attempt at a Solution


may can i have some ideas how to find this? i tried to find the x but i don't know how with the roots.
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 .
 
  • #3
SammyS
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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]​
 
  • #4
HallsofIvy
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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.
 

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