Radicals with order higher than 2

[Eurovision]

Homework Statement

order 3 radical from (9+4radical5) + order 3 radical from (9-4radical5)

*note: 9+4radical5 is all under the big order 3 radical. (so is the other one)

Homework Equations

in case you don't understand what i wrote (sorry, my language is not english and math linguistics differ in my language) click the attachement to see a picture of the equation

The Attempt at a Solution

I think i might find the solution by writing 9+4radical5 as a cube of something so as to cancel the radical but what if i have order 6 radical? (because i have this type of exercises as the next one. So clearly writing as a square, cube etc won't work everytime. Thank you!

 

[SammyS]

Homework Statement

order 3 radical from (9+4radical5) + order 3 radical from (9-4radical5)

*note: 9+4radical5 is all under the big order 3 radical. (so is the other one)

Homework Equations

in case you don't understand what i wrote (sorry, my language is not english and math linguistics differ in my language) click the attachement to see a picture of the equation

The Attempt at a Solution

I think i might find the solution by writing 9+4radical5 as a cube of something so as to cancel the radical but what if i have order 6 radical? (because i have this type of exercises as the next one. So clearly writing as a square, cube etc won't work everytime. Thank you!

Hallo Eurovision. Welcome to PF !

I take it that you need to in some way simplify $\displaystyle \sqrt[3]{9+4\sqrt{5}\ }+\sqrt[3]{9-4\sqrt{5}\ }\,.$ Is that correct?

 

[Eurovision]

Thank you.

Yes, i must find what it equals to (I know already is 3 from the answers) but i don't know how to solve it.

 

[Infinitum]

Try using the binomial theorem for index n=1/3.

 

[HallsofIvy]

I would do this is in a rather different way. Cardano's formula for the "reduced" cubic, $x^3+ px+ q= 0$, gives a result of the form
$$x= \sqrt[3]{-\frac{q}{2}+ \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+ \sqrt[3]{-\frac{q}{2}- \sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}}$$

The number in this problem, being a sum of cube roots, is of that form with -q/2= 9 and $\sqrt{q^2/4+p^3/27}= 4\sqrt{5}$ so that $q^2/4+ p^3/27= 80$. Solving for p and q gives you a cubic equation that has x= 3 as its only real root.

 

[Infinitum]

I would do this is in a rather different way. Cardano's formula for the "reduced" cubic, $x^3+ px+ q= 0$, gives a result of the form
$$x= \sqrt[3]{-\frac{q}{2}+ \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+ \sqrt[3]{-\frac{q}{2}- \sqrt{\frac{q^2}{4}+ \frac{p^3}{27}}}$$

The number in this problem, being a sum of cube roots, is of that form with -q/2= 9 and $\sqrt{q^2/4+p^3/27}= 4\sqrt{5}$ so that $q^2/4+ p^3/27= 80$. Solving for p and q gives you a cubic equation that has x= 3 as its only real root.

Interesting method. *like*

 

[professordad]

Here's how I would do it: cube the entire thing.

Something helpful: If you have $S = A + B$, then $S^3 = A^3 + B^3 + 3AB(A + B)$. And what is $A + B$? :)

 

[SammyS]

Notice that $\displaystyle \sqrt[3]{9+4\sqrt{5}\ }+\sqrt[3]{9-4\sqrt{5}\ }\,$


can be written as $\displaystyle \sqrt[3]{\sqrt{81}+\sqrt{80}\ }+\sqrt[3]{\sqrt{81}-\sqrt{80}}\ .$