Prove the following:
-Dan
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
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?