# Thread: Simplify this expression...

1. ## Simplify this expression...

$\displaystyle ({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}$

2. Originally Posted by Alex2103
$\displaystyle ({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}$
Get ready for a ride....

Here are the highlights:
Let $\displaystyle \displaystyle x = ( 2 + \sqrt{5} )^{1/3} + (2 - \sqrt{5} )^{1/3}$

Now
$\displaystyle \displaystyle x^3 = (2 + \sqrt{5}) + 3( 2 + \sqrt{5} )^{2/3}(2 - \sqrt{5} )^{1/3} + 3(2 - \sqrt{5} )^{1/3}(2 - \sqrt{5})^{2/3} + (2 - \sqrt{5})$

This reduces to
$\displaystyle \displaystyle x^3 = 4 -3(2 + \sqrt{5})^{1/3} - 3(2 - \sqrt{5})^{1/3}$

or, reminding ourselves of what x is equal to:
$\displaystyle \displaystyle x^3 = 4 -3x$

Now solve:
$\displaystyle \displaystyle x^3 + 3x - 4 = 0$

(Hint: There is only one real solution.)

-Dan

3. Hello, Alex2103!

$\displaystyle \text{Simplfy: }\;\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}$

A binomial of the form $\displaystyle a + b\sqrt{5}$ makes me suspect
. . that the problem involves the Golden Mean: .$\displaystyle \phi \:=\:\dfrac{1+\sqrt{5}}{2}$

$\displaystyle \text{And sure enough: }\;\phi^3 \;=\;\left(\frac{1+\sqrt{5}}{2}\right)^3 \;=\;2+\sqrt{5}$

. . . $\displaystyle \text{and we find that: }\;\left(\frac{1-\sqrt{5}}{2}\right)^3 \;=\;2-\sqrt{5}$

Therefore:

. . $\displaystyle \sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} \;\;=\;\;\sqrt[3]{\left(\frac{1+\sqrt{5}}{2}\right)^3} + \sqrt[3]{\left(\frac{1-\sqrt{5}}{2}\right)^3}$

. . $\displaystyle =\;\;\dfrac{1 + \sqrt{5}}{2} + \dfrac{1-\sqrt{5}}{2} \;\;=\;\;1$

I love your solution, Dan!

4. Originally Posted by Soroban
I love your solution, Dan!
But yours is so more elegant and less time consuming!

-Dan

5. Thanks for help!

6. Heres' yet another way to do that. Cardano's formula for the reduced cubic equation says that a solution to $\displaystyle x^3+ mx= n$ is of the form $\displaystyle \sqrt[3]{\frac{n}{2}+ \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}+ \sqrt[3]{\frac{n}{2}- \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}$.

Looks familiar, doesn't it? That is precisely the same as $\displaystyle \sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}$ with $\displaystyle \frac{n}{2}= 2$ and $\displaystyle \frac{n^2}{4}+ \frac{m^3}{27}= 5$.

From $\displaystyle \frac{n}{2}= 2$, n= 4 and then $\displaystyle \frac{n^2}{4}+ \frac{m^3}{27}= 4+ \frac{m^3}{27}= 5$ so that $\displaystyle \frac{m^3}{27}= 1$, $\displaystyle m^3= 27$, and $\displaystyle m= 3$.

That means that this number is a real root of $\displaystyle x^3+ 3x= 4$ or $\displaystyle x^3- 3x- 4= 0$. It is easy to see that $\displaystyle x^3+ 3x- 4= (x- 1)(x^2+ x+ 4)$. Since the discriminant of $\displaystyle x^2+ x+ 4$ is $\displaystyle 1- 16= -15$ the only real root of that equation is 1 so we must have $\displaystyle \sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}= 1$.