Problem algebra involving third roots

[ParisSpart]

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 don't know how with the roots..

 

[SammyS]

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

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

(a - b)³ = a³ - 3(a²)b + 3a b² - b³ .

 

[SammyS]

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

Also it may be helpful to express, (a-b)^3 as

a^3-b^3-3ab(a-b)\ .​

 

[HallsofIvy]

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

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 x^3+ mx= n. From m= 3ab, b= m/(3a) and then n= a^3- m^3/(3^3a^3). Multiplying by a^3 gives na^3= (a^3)^2- (m/3)^3] which is a quadratic in a^3, (a^3)^2- na^3- (m/3)^3= 0, which can be solved by the quadratic equation:
a^3= \frac{n\pm\sqrt{n^2+ 4(m/3)^3}}{2}= \frac{n}{2}\pm\sqrt{(n/2)^2+ (m3)^3}

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

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 (n/2)^2- (m/3)^3= 100- (m/3)^3= 108 so that (m/3)^3= -8, m/3= -2, and m= -6.