For this solution, I'll be using the same technique that Youtuber blackpenredpen used in this video.
To compute this expression, let it equal x:
x=root3(2+sqrt5)+root3(2-sqrt5)
Now, cube both sides using the expansion (a+b)^3=a^3+3a^2b+3ab^2+b^3 (it's gonna get ugly, but just trust me here):
x^3=(root3(2+sqrt5)+root3(2-sqrt5))^3
x^3=(root3(2+sqrt5))^3 + 3(root3(2+sqrt5))^2(root3(2-sqrt5)) + 3(root3(2+sqrt5))(root3(2-sqrt5))^2+(root3(2-sqrt5))^3
Simplify the cube roots and the exponents:
x^3=2+sqrt5 + 3(root3(2+sqrt5))^2(root3(2-sqrt5)) + 3(root3(2+sqrt5))(root3(2-sqrt5))^2+2-sqrt5
Collect like terms:
x^3=4 + 3(root3(2+sqrt5))^2(root3(2-sqrt5)) + 3(root3(2+sqrt5))(root3(2-sqrt5))^2
Now, bring the cube roots that are bring multiplied under one radical, like this:
x^3=4 + 3root3((2+sqrt5)^2)*root3(2-sqrt5) + 3root3(2+sqrt5)*root3((2-sqrt5)^2)
x^3=4 + 3root3((2+sqrt5)^2(2-sqrt5)) + 3root3((2+sqrt5)(2-sqrt5)^2)
Rewrite the squared terms, then use the difference of squares factoring:
x^3=4 + 3root3((2+sqrt5)(2+sqrt5)(2-sqrt5)) + 3root3((2+sqrt5)(2-sqrt5)(2-sqrt5))
x^3=4 + 3root3((2+sqrt5)(4-5)) + 3root3((4-5)(2-sqrt5))
x^3=4 + 3root3(-(2+sqrt5)) + 3root3(-(2-sqrt5))
The cube root of -1 is -1:
x^3=4 - 3root3(2+sqrt5) - 3root3(2-sqrt5)
x^3=4 - 3(root3(2+sqrt5) +root3(2-sqrt5))
This is the original x:
x^3=4 - 3x
x^3+3x-4=0
Use the rational roots theorem and synthetic division to figure out that 1 is the only rational root of the polynomial:
https://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.php
So the polynomial can be factored as:
(x-1)(x^2+x+4)=0
Since x^2+x+4 has no real solutions, the only solution is:
x=1
That's the solution (finally). Hope this helped!