A Nice Algebra Simplification | Math Olympiad | Give It a Try! 

Let t=2^1/6. Then, x=1+t+t^2+t^3 +t^4 +t^5 = [t^6-1]/(t-1) = 1/(t-1). So x+1 = t/(t-1). Now, 1/x+2/x^2+1/x^3 = 1/x^3(x+1)^2 = [t^2(t-1)^3]/(t-1)^2 = t^2(t-1) = t^3-t^2 = 2^(1/2) - 2^(1/3).

Isn't x the sum of a finite geometric series?

We have, (1/x) + (2/x²) + (1/x³) = (x² + 2x + 1) /x³ = (x + 1)² /x³ x = 1 / { (2)⅙ - 1 } x + 1 = [ 1 / { (2)⅙ - 1 } ] + [ { (2)⅙ - 1 } / { (2)⅙ - 1 } ] = (2)⅙ / { (2)⅙ - 1 } And, 1 /x = (2)⅙ - 1 Therefore, (x + 1)² /x³ = [ (2)⅙ / { (2)⅙ - 1 } ]² { (2)⅙ - 1 }³ = (2)⅓ { (2)⅙ - 1 } = (2)⅓(2)⅙ - (2)⅓ = (2)½ - (2)⅓ = √2 - (2)⅓

Evaluating x using Method 2 is more direct. From 10:51, let a = 2^(1/6) for simplicity and let u = 1/x = a - 1. (Don't bother to simplify x by rationalizing the denominator.). Then, note that the desired expression can be written as: u(1 + u)^2 = (a - 1)(a^2) = [2^(1/6) - 1][2^(2/6)] = 2^(3/6) - 2^(2/6) = sqrt2 - 2^(1/3)

(2^(1/3)/2^(1/6) -1)

let a=2^(1/6), now you have obtained (1/x)=(a-1) ==>(1/x) +(2/x^2)+(1/x^3) = a^3-a^2 = 2^(1/2)-2^(1/3). You seemed to have taken an unnecessary detour once (1/x) was determined

E=√2-³√2

2^(1/2)-2^(1/3). χ=(1-α^6)/(1-α) οπου α=2^(1/6) Ε=(1/χ)(1+1/χ)^2

{2x^2+24x^2}=26x^4 48x^2 {26x^4+48x^2}=74x^6 96x^2 {74x^6+96x^2}= 170x8 124x^2 {170x^8+124x^2}= 294x^10 10^20 2^37 x^2^5 2^55^4 2^37^1x^2^5 1^1^11^2^2^2^1^1x^2^1 1^1^1^1x^2^1 x^2^1 (x ➖ 2x+1) 1x+1x ➖/x+x ➖ +2x+2x ➖/x^2+x ^2➖ =4x+4x ➖/x^2+x^2 ➖ 2x^2/x^2+4x^2/x^4+8x^2/x^4 14x^10/x^10.=:1.4x^1 1^1.2^2x1^1 1^2x^1^1 2x^1 (x ➖ 2x+1)