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

1 Answer
Apr 28, 2018

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:

https://www.mathportal.org/calculators/polynomials-solvers/synthetic-division-calculator.phphttps://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!