Question

How to solve `root(3)(2+sqrt(5))+root(3)(2-sqrt(5))`?

Answer 1

The result is `1`.

Explanation:

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!