For this solution, I'll be using the same technique that Youtuber blackpenredpen used in this video.
To compute this expression, let it equal xx:
x=root3(2+sqrt5)+root3(2-sqrt5)x=3√2+√5+3√2−√5
Now, cube both sides using the expansion (a+b)^3=a^3+3a^2b+3ab^2+b^3(a+b)3=a3+3a2b+3ab2+b3 (it's gonna get ugly, but just trust me here):
x^3=(root3(2+sqrt5)+root3(2-sqrt5))^3x3=(3√2+√5+3√2−√5)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))^3x3=(3√2+√5)3+3(3√2+√5)2(3√2−√5)+3(3√2+√5)(3√2−√5)2+(3√2−√5)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-sqrt5x3=2+√5+3(3√2+√5)2(3√2−√5)+3(3√2+√5)(3√2−√5)2+2−√5
Collect like terms:
x^3=4 + 3(root3(2+sqrt5))^2(root3(2-sqrt5)) + 3(root3(2+sqrt5))(root3(2-sqrt5))^2x3=4+3(3√2+√5)2(3√2−√5)+3(3√2+√5)(3√2−√5)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)x3=4+33√(2+√5)2⋅3√2−√5+33√2+√5⋅3√(2−√5)2
x^3=4 + 3root3((2+sqrt5)^2(2-sqrt5)) + 3root3((2+sqrt5)(2-sqrt5)^2)x3=4+33√(2+√5)2(2−√5)+33√(2+√5)(2−√5)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))x3=4+33√(2+√5)(2+√5)(2−√5)+33√(2+√5)(2−√5)(2−√5)
x^3=4 + 3root3((2+sqrt5)(4-5)) + 3root3((4-5)(2-sqrt5))x3=4+33√(2+√5)(4−5)+33√(4−5)(2−√5)
x^3=4 + 3root3(-(2+sqrt5)) + 3root3(-(2-sqrt5))x3=4+33√−(2+√5)+33√−(2−√5)
The cube root of -1−1 is -1−1:
x^3=4 - 3root3(2+sqrt5) - 3root3(2-sqrt5)x3=4−33√2+√5−33√2−√5
x^3=4 - 3(root3(2+sqrt5) +root3(2-sqrt5))x3=4−3(3√2+√5+3√2−√5)
This is the original xx:
x^3=4 - 3xx3=4−3x
x^3+3x-4=0x3+3x−4=0
Use the rational roots theorem and synthetic division to figure out that 11 is the only rational root of the polynomial:
So the polynomial can be factored as:
(x-1)(x^2+x+4)=0(x−1)(x2+x+4)=0
Since x^2+x+4x2+x+4 has no real solutions, the only solution is:
x=1x=1
That's the solution (finally). Hope this helped!
