Prove the following:

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

-Dan

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- Apr 27th 2014, 04:03 AMtopsquarkProblem #3 - Cubic Expression
Prove the following:

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

-Dan - Apr 27th 2014, 06:25 AMprasumre: 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 - Apr 27th 2014, 08:33 AMromsekre: Problem #3 - Cubic Expression
- Apr 27th 2014, 09:06 AMSorobanre: Problem #3 - Cubic Expression
Hello, topsquark!

Quote:

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

__Spoiler__:

- Apr 27th 2014, 09:53 AMprasumre: Problem #3 - Cubic Expression
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__Spoiler__: - May 1st 2014, 12:39 PMJeffMre: 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? - May 2nd 2014, 04:14 AMprasumre: 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 - May 3rd 2014, 12:51 AMtopsquarkRe: Problem #3 - Cubic Expression
Congrats to everyone this week! (Clapping)

-Dan