Show Radical Equation is Equal

Show that both sides of radical equation is equal. If it is not equal, explain why.

Let cbrt = cube root for short

Let sqrt = square root for short


cbrt(2 + sqrt{5}) + cbrt(2 - sqrt{5}) = 1

Let x=cbrt(2 + sqrt{5}), y=cbrt(2 - sqrt{5}), and A=x+y.

A^3=x^3 + y^3 +3x^2 y + 3xy^2 =

2+sqrt(5) + 2-sqrt(5) +3xy(x+y) =

4+ 3cbrt[(2+sqrt(5)*(2-sqrt(5) ]*A (because A=x+y)

= 4+3cbrt(-1)*A =4-3A

So A^3=4-3A.

So A^3+3A-4=0

1 is a solution for this equation. So there exist b,c such that A^3+3A-4=(A-1)(A^2+bA+c)

So A^3+3A-4=A^3+bA^2+cA-A^2-bA-c

So b=1 and c=4.

A^3-3A+4=(A-1)(A^2+A+4) =0

For the equation A^2+A+4=0, delta=1-4*4*1 <0. Then, since (A-1)(A^2+A+4) =0 and A^2+A+4 could not be equal to 0, we conclude that A=1. :)