# Thread: Problem #3 - Cubic Expression

1. ## Problem #3 - Cubic Expression

Prove the following:

$\displaystyle \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} = 3$

-Dan

2. ## re: Problem #3 - Cubic Expression

easy let cuberoot(9+sqrt(80))+ cuberoot(9-sqrt(80))=x
now cube on both l.h.s and r.h.s
apply (a+b)^(3)=a^3+b^3+3ab(a+b)
now
x^(3)=9+sqrt(80)+9-sqrt(80)+3*cuberoot(9+sqrt(80))*cuberoot(9-sqrt(80))*(x)
now (9-sqrt(80))=1/(9+sqrt(80))
therefore the two cuberoot terms would be cancelled
now
x^(3)=18-3x
x^(3)+3x-18=0
now apply factor theorem take (x-3) and divide this to the above cubic eqn
we thus get x=3 as root
substitute the value of x and u will get the result

3. ## re: Problem #3 - Cubic Expression

Originally Posted by prasum
easy <snip>

Spoiler:
like this

Others would like a chance to do the problem themselves.

It's the button that looks like an eye on the advanced editing toolbar.

4. ## re: Problem #3 - Cubic Expression

Hello, topsquark!

Prove: .$\displaystyle \sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} \;=\: 3$

Spoiler:

Let: .$\displaystyle A \,=\,9 + \sqrt{80},\;B \,=\,9-\sqrt{80}$

Note that: .$\displaystyle A + B \:=\: (9+\sqrt{80}) + (9-\sqrt{80}) \:=\:18$
. . . . . . . . . . . $\displaystyle AB \:=\:(9+\sqrt{80})(9 - \sqrt{80}) \:=\:1$

We have: .$\displaystyle X \:=\:A^{\frac{1}{3}} + B^{\frac{1}{3}}$

Cube both sides:
. . $\displaystyle X^3 \;=\;\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)^3$

. . $\displaystyle X^3 \;=\;A + 3A^{\frac{2}{3}}B^{\frac{1}{3}} + 3A^{\frac{1}{3}}B^{\frac{2}{3}} + B$

. . $\displaystyle X^3 \;=\;(A + B) + 3A^{\frac{1}{3}}B^{\frac{1}{3}}\left(A^{\frac{1}{3 }} + B^{\frac{1}{3}}\right)$

. . $\displaystyle X^3 \;=\;\underbrace{(A+B)}_{18} + 3\underbrace{(AB)^{\frac{1}{3}}}_{1^\frac{1}{3}}} \underbrace{\left(A^{\frac{1}{3}} + B^{\frac{1}{3}}\right)}_3$

. . $\displaystyle X^3 \;=\;18 + 3(1)(3)$

. . $\displaystyle X^3 \;=\;27$

Therefore: .$\displaystyle X \;=\;3$

5. ## re: Problem #3 - Cubic Expression

.
Spoiler:
easy let cuberoot(9+sqrt(80))+ cuberoot(9-sqrt(80))=x
now cube on both l.h.s and r.h.s
apply (a+b)^(3)=a^3+b^3+3ab(a+b)
now
x^(3)=9+sqrt(80)+9-sqrt(80)+3*cuberoot(9+sqrt(80))*cuberoot(9-sqrt(80))*(x)
now (9-sqrt(80))=1/(9+sqrt(80))
therefore the two cuberoot terms would be cancelled
now
x^(3)=18-3x
x^(3)+3x-18=0
now apply factor theorem take (x-3) and divide this to the above cubic eqn
we thus get x=3 as root
substitute the value of x and u will get the result

6. ## re: Problem #3 - Cubic Expression

I must admit that it took me a while to understand prasum's answer, but I eventually got there. Possibly I drank too much wine with dinner last night because I still don't understand soroban's answer.

Prove $\sqrt[3]{9 + \sqrt{80}} + \sqrt[3]{9 - \sqrt{80}} = 3.$

Eventually prasum gets to $x^3 = 18 - 3x$ but he means $x^3 = 18 + 3x.$

$x^3 = 18 + 3x \implies x^3 - 3x - 18 = 0 \implies (x - 3)(x^2 + 3x + 6) = 0 \implies x = 3\ if\ x \in \mathbb R.$

So the typo had me stuck for a while.

Soroban defines $A = 9 + \sqrt{80}\ and\ B = 9 - \sqrt{80}.$ In his notation he must prove $\sqrt[3]{A} + \sqrt[3]{B} = 3.$ I'm fine with that obviously.

He asserts that $A + B = 18.$ Even I can deduce that from his definitions.

He asserts that $\sqrt[3]{AB} = 1.$ I can also deduce that from his definitions.

And he asserts that $\sqrt[3]{A} + \sqrt[3]{B} = 3.$ But he can't assert that because that is what is to be proved, or am I missing something?

7. ## re: Problem #3 - Cubic Expression

yes the secondlast step of Sorobans answer must have (A^(1/3)+B^(1/3))=X not 3
then X would be the required answer

8. ## Re: Problem #3 - Cubic Expression

Congrats to everyone this week!

-Dan