cubed root (9+4 sqrt 5)+cubed root (9-4 sqrt 5) = 3

CPhill's answer is, in the words of one of my instructors, elegant.

That won't be said for mine; more like a brute force attempt:

To make the typing easier,

let  A  =  (9 + 4√5)^(1/3)          --->          A³  =  9 + 4√5

let  B  =  (9 - 4√5)^(1/3)           --->          B³  =  9 - 4√5

Let  X  =  A + B             So, now:  X  = (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3)    (the problem)

Also:   A·B  =  [ (9 + 4√5)^(1/3) ] · [ (9 - 4√5)^(1/3) ]

                 =  [ (9 + 4√5) · (9 - 4√5) ] ^(1/3)

                 =  [ 81 - 16 · 5 ] ^(1/3)

                 =  [ 1 ] ^(1/3)

                 =  1

Since  A·B  =  1,

     A²·B  =  A(AB)  =  A(1)  =  A  =  (9 + 4√5)^(1/3) 

     A·B²  =  (AB)B  =  (1)B  =  B  =  (9 - 4√5)^(1/3) 

Sinc  X  =  A + B,

        X³  =  (A + B)³  =  A³ + 3A²·B + 3A·B² + B³

        X³  =  (9 + 4√5) + [ 3(9 + 4√5)^(1/3) ] + [ 3(9 - 4√5)^(1/3) ] + ( 9 - 4√5 )

Rearranging:

        X³  =  18 + 3[ (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) ]

        X³  =  18 + 3[ A + B ]

But, since A + B  =  X

       X³  =  18 + X

       X³ - X - 18  =  0

Factoring:

      (X - 3)(X² + 3X + 6)  = 0

So:  X = 3  or X = [-3 ± i√(15) ] / 2

Since the answer is a pure real number, the answer is  3!

$$\sqrt[3]{(9+4 \sqrt 5}+\sqrt[3]{(9-4 \sqrt 5} = 3$$

Umm I do not know - I'll have to think about this one!

This one is tricky.....but note.....(9 + 4√5)^(1/3) = (Phi)²  where Phi = (1 + √5)/2

And (9 -4√5)^(1/3) = phi² = Phi ^-2 where Phi ^-1  = phi = 2/(1 + √5)

So we have

Phi ² + Phi ^-2   =

Phi ² + phi ²       and by a property of Phi and phi......Phi² = 1 + Phi   and phi² = 1 - phi ... so we have

(1 + Phi)  + (1 - phi) =

2 + Phi - phi            and by another property of Phi and phi.....Phi - phi = 1    so we have

2 + 1 = 3

....but note.....(9 + √80)^(1/3) = (Phi)2  where Phi = (1 + √5)/2 

Christ Chris, What archive did you dig that out of !

When I evaluated (9  + √80)^(1/3) and (9 - √80)^(1/3), I immediately saw that these were related to Phi and phi.....

Thanks Chris and Gino :)))

Very nice, Geno.....I would dispute that mine is more "elegant"....more like...."calculator aware".... LOL!!!

(Plus.....a little knowledge about the proprties of Phi and phi....!!)

Yea well - it is all still too much for me LOL  :)))