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🧵 How do we actually prove this?

Anonymous No. 16586665

The calculator says that this radical sum equals 1 but I can't find any proof. Can you help?

Anonymous No. 16586686

>>16586665
Here's a hint: [math]1^{3}=1=1^{2}[/math]

Anonymous No. 16586700

>>16586665
wolframalpha

Anonymous No. 16586997

>>16586665
it's false
proof:
https://www.wolframalpha.com/input?i=%282+%2B+sqrt%285%29%29%5E%281%2F3%29+%2B+%282+-+sqrt%285%29%29%5E%281%2F3%29+%3D+1

Anonymous No. 16587012

>>16586997
>Assuming the principal root | Use the real‐valued root instead
Wtf? I you click "Use the real‐valued root instead", suddenly it is true instead. What does Stephen Wolfram meant by this?

Anonymous No. 16587014

>>16586665
I just plugged it into a calculator and got 1.059....
I think the calculator result would be much closer to 1 if your statement were true.

Anonymous No. 16587057

>>16586665
let x:= the left side of this identity. Then (by using the formula (a+b)^3 = a^3 + b^3 + 3ab (a+b) valid for every real numbers a,b: notice that in this specific example we have also ab = -1) we have also the equality x^3 = 4 - 3x. Hence x^3 +3x - 4 = 0. This polynomial has 1 as an obvious root, hene it can be divided by (x-1). if you compute the euclidean division by (x-1) you find the factorization x^3 + 3x - 4 = (x^2 + x + 4) * (x - 1). But the discriminant of x^2 + x +4 is negative (this polynomial turns out to be (x + 1/2)^2 + 3/4) and so x^2 +x + 4 has no real roots and thus 1 is he only real root of (x^2+x+4)(x-1) = x^2 + 3x - 4. Thus 1 is the quantity in the pic.

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Anonymous No. 16587089

>>16587012
the OP tricked me
the depicted equation is true
proof:
https://www.wolframalpha.com/input?i=%28sqrt%285%29+%2B+2%29%5E%281%2F3%29+-+%28sqrt%285%29+-+2%29%5E%281%2F3%29+%3D+1

Anonymous No. 16587140

>>16586665

s = sqrt(5)

a = (s + 2)^(1/3) = (s + 1)/2

b = (s – 2)^(1/3) = (s – 1)/2

a – b
= (s + 1)/2 – (s – 1)/2
= (s + 1 – s + 1)/2
= 2/2
= 1

Anonymous No. 16587502

>>16586665
Construct it with straight edge and compass