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.