# Simplify this expression...

• March 5th 2011, 02:15 PM
Alex2103
Simplify this expression...
$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$
• March 5th 2011, 03:33 PM
topsquark
Quote:

Originally Posted by Alex2103
$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$

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

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

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

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

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

(Hint: There is only one real solution.)

-Dan
• March 5th 2011, 03:55 PM
Soroban
Hello, Alex2103!

Quote:

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

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

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

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

Therefore:

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

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

• March 5th 2011, 04:07 PM
topsquark
Quote:

Originally Posted by Soroban

But yours is so more elegant and less time consuming! (Bow)

-Dan
• March 6th 2011, 04:51 AM
Alex2103
Thanks for help!(Happy)
• March 6th 2011, 12:05 PM
HallsofIvy
Heres' yet another way to do that. Cardano's formula for the reduced cubic equation says that a solution to $x^3+ mx= n$ is of the form $\sqrt[3]{\frac{n}{2}+ \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}+ \sqrt[3]{\frac{n}{2}- \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}$.

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

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

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