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:
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!
