Recognize this as an inverse golden ratio problem and solve it easily. Let φ=(√5-1)/2 be the solution of φ^2+φ-1=0 or φ^2=1-φ. Then φ^3=φ-φ^2=φ-(1-φ)=2φ-1=2(√5-1)/2-1=√5-2. Thus, ∛(√5-2)=∛(φ^3)=φ=(√5-1)/2
Suppose -2+√5 = (u +√v)³ where u & v are rational & v is not the square of a rational. (u² - v)³ = (u + √v)³.(u - √v)³ = (- 2 + √5).(- 2 - √5) = 4 - 5 = - 1 = (-1)³, so u and v exists. If we did not get the cube of a rational number, then u and v would not have existed. Now - 2 + √5 = (u³ + 3uv) + (3u² + v)√v. So - 2 = (u³ + 3uv) and √5 = (3u² + v)√v. Therefore (u-√v)³ = (u³+3uv) - (3u²+v)√v = - 2 - √5. So - 1 = 4 - 5 = (- 2 +√5).(- 2 -√5) = [(u + √v)³].[(u - √v)³] = (u² - v)³. Hence (u² - v) = ∛(-1) = - 1 & so v = u² - (-1) = u² +1. Therefore - 2 = (u³ + 3u.v) = u³ + 3u.(u² + 1) = 4.u³ + 3u. So 4.u³ + 3.u + 2 = 0. By the Rational Root theorem, u = -½, bec. 4u³ + 3u + 2 = (2u+1).(2u² - u +2) & the quadratic part has no real roots. ∴ v = u²+1 = ⁵/₄ . So ∛(-2 +√5) = (u+√v) = -½ + √(⁵/₄) = (√5 -1)/2. .
Recognize this as an inverse golden ratio problem and solve it easily. Let φ=(√5-1)/2 be the solution of φ^2+φ-1=0 or φ^2=1-φ. Then φ^3=φ-φ^2=φ-(1-φ)=2φ-1=2(√5-1)/2-1=√5-2. Thus, ∛(√5-2)=∛(φ^3)=φ=(√5-1)/2
Suppose -2+√5 = (u +√v)³ where u & v are rational & v is not the square of a rational.
(u² - v)³ = (u + √v)³.(u - √v)³ = (- 2 + √5).(- 2 - √5) = 4 - 5 = - 1 = (-1)³, so u and v exists.
If we did not get the cube of a rational number, then u and v would not have existed.
Now - 2 + √5 = (u³ + 3uv) + (3u² + v)√v. So - 2 = (u³ + 3uv) and √5 = (3u² + v)√v.
Therefore (u-√v)³ = (u³+3uv) - (3u²+v)√v = - 2 - √5. So - 1 = 4 - 5 = (- 2 +√5).(- 2 -√5)
= [(u + √v)³].[(u - √v)³] = (u² - v)³. Hence (u² - v) = ∛(-1) = - 1 & so v = u² - (-1) = u² +1.
Therefore - 2 = (u³ + 3u.v) = u³ + 3u.(u² + 1) = 4.u³ + 3u. So 4.u³ + 3.u + 2 = 0. By the
Rational Root theorem, u = -½, bec. 4u³ + 3u + 2 = (2u+1).(2u² - u +2) & the quadratic
part has no real roots. ∴ v = u²+1 = ⁵/₄ . So ∛(-2 +√5) = (u+√v) = -½ + √(⁵/₄) = (√5 -1)/2.
.