Prove that (√5 + 2)^(1/3) - (√5 - 2)^(1/3) = 1?

Question

Some of you seem to have a knack for this kind of puzzle. Try this one.

Answer 1

Let n=(√5 + 2)^(1/3) - (√5 - 2)^(1/3)
n^3=(√5+2)-3((√5+2)^1/3)+3((√5-2)^1/3)-(√5-2)=4-3n
n^3+3n-4=0
(n-1)(n^2+n+4)=0
So n=1

Answer 2

(√5 + 2)^(1/3) - (√5 - 2)^(1/3) = 1

I had a proof but got stuck. But I'm going to try again.
Let x = √5 + 2 and y = √5 - 2.

LHS = x^(1/3) - y^(1/3)

I'm going to rationalize this using the fact that
(a^3 - b^3) = (a - b)(a^2 + ab + b^2)
It follows that
(a - b) = (a^(1/3) - b^(1/3)) (a^(2/3) + (ab)^(1/3) + b^(2/3))

LHS = (x - y) / ( x^(2/3) + (xy)^(1/3) + y^(2/3) )

xy = (√5 + 2) (√5 - 2) = 5 - 4 = 1

LHS = (x - y) / ( x^(2/3) + (1)^(1/3) + y^(2/3) )

LHS = (x - y) / ( x^(2/3) + 1 + y^(2/3) )

We can express x^(2/3) as (x^2)^(1/3). Similarly with y^(2/3).

LHS = (x - y) / ( (x^2)^(1/3) + 1 + (y^2)^(1/3) )

x^2 = [√5 + 2]^2 = 5 + 4√5 + 4 = 9 + 4√5
y^2 = [√5 - 2]^2 = 5 - 4√5 + 4 = 9 - 4√5

x - y = (√5 + 2) - (√5 - 2) = 4

LHS = 4 / ( (9 + 4√5)^(1/3) + 1 + (9 - 4√5)^(1/3) )

What choice do we have but to rationalize the numerator within the first radical?

9 + 4√5 = [(9 + 4√5)(9 - 4√5)] / (9 - 4√5)
= [81 - 80] / (9 - 4√5)
= 1 / (9 - 4√5)

LHS = 4 / ( (1 / (9 - 4√5))^(1/3) + 1 + (9 - 4√5)^(1/3) )

LHS = 4 / ( 1 / (9 - 4√5)^(1/3) + 1 + (9 - 4√5)^(1/3) )

Multiplying top and bottom by (9 - 4√5).

LHS = 4(9 - 4√5) / [ 1 + (9 - 4√5)^(1/3) + (9 - 4√5)^(2/3) ]

I can't get anywhere. But I thank you though for presenting this problem (as I like puzzles). I can probably try other stuff but am too tired at the moment.

Answer 3

Let x = (sqrt(5) + 2)^(1/3) and y = (sqrt(5) - 2)^(1/3). Then:

4 = x^3 - y^3 = (x-y)*(x^2 +y^2 +xy) = (x-y)*((x-y)^2 +3xy).

Note that xy = (5-4)^(1/3) = 1, so

4 = (x-y)*((x-y)^2 + 3).

Now let z = x-y. Then:

4 = z^3 + 3 z

so
0 = z^3 + 3z -4 = (z-1)(z^2 +z +4).

Since z = x-y = (√5 + 2)^(1/3) - (√5 - 2)^(1/3) must be real, and z^2 + z + 4 is irreducible over the real numbers, z = 1. This is the desired result.

Answer 4

let a^3 = (2+sqrt(5))
b^3 = 2- sqrt(5)
c^3 = -1
we have a^3+b^3+c^3 = +3

3 abc = 3(2+sqrt(5)(2-sqrt(5))(-1) =3(4-5)(-1) = 3

so a^3+b^3+c^3 = 3bc

so a+b+c = 0

so (2+5^(1/2))^(1/3) + (2-sqrt(5))^(1/3) - 1 = 0
rearanging we get

(2+5^(1/2))^(1/3)-(5^(1/3)-2)^(1/3) = 1

hence proved

Answer 5

The (square root of 5) + 2 = 4.236067977... and the cube root of that number (that number to the (1/3) power) is 1.6180339887...
The (square root of 5) - 2 = .2360679775... and the cube root of that number (that number to the (1/3) power) is .6180339887...
When you subtract the first one from the second (1.6180339887 - .6180339887) it turns out to be 1.

Answer 6

yay i solved it!!! ok here's wut i did uhm...except i don't know how you typed the square root sign so imma just write squareroot for the square root sign ok?

squareroot(5) + 2 = 4.236067977
4.236067977^(1/3) = 1.618033989 <-----

now squareroot(5) - 2 = .2360679775
.2360679775^(1/3) = .6180339887<----

so taking the two answers with the arrows next to them...and subtract them...u get

1.618033989 - .6180339887 = 1