Amazing Algebra Problem | Can You Try?

[@Shobhamaths]

E=89√3👍

[@RashmiRay-c1y]

Let y=2x. Note that 2E= 32x^5 + 1/(32x^5) = y^5+1/y^5. Now x= 1/2(√3 - √2). Thus, y=√3 - √2 and 1/y = √3 + √2. So y+1/y = 2√3, y^2+1/y^2= 10, y^3+1/y^3 = 18√3. So, (y^2+1/y^2)(y^3+1/y^3)= 180√3 = 2E + y + 1/y = 2E + 2√3. Therefore E =(178/2)√3 = 89√3.

[@gregevgeni1864]

Κ = 89√3 .

[@RealQinnMalloryu4]

{1+1 ➖ }/4+{x+x ➖ }={2/4+x^2 }=2x^2/4 1x^1/2^2 x^1/1^2 (x ➖ 2x+1).6/{2+2 ➖}= 6/4=1.2 (x ➖ 2x+1).18+{x^5+x^5 ➖ }{1+1 ➖ }/16+{x^5+x^5 ➖ }={8+x^10}+2/{16+x^10}={8x^10+2}/16x^10=10x^10/16x^10 5^5x^5^5/8^8x^5^5 2^3^2^3x^2^3^2^3/2^3^2^3x^2^3^2^3 1^1^1^1x^1^1^1^1/1^1^1^1^1^1x^2^1^1^3 /x^2^3(x ➖ 3x+2).

[@Fjfurufjdfjd]

Ε=16χ^5+1/(64χ^5) 2Ε=32χ^5+2/(64χ^5)
2Ε=(2χ)^5+1/(2χ)^5 σχεση 1.
χ=[(3)^(1/2)-(2)^(1/2)]/2 2χ=(3)^(1/2)-(2)^(1/2)
1/2χ=(3)^(1/2)+(2)^(1/2) αρα
2χ+1/2χ=2(3)^(1/2)
2Ε=[2(3)^(1/2)]^5-5×18(3)^(1/2)-20(3)^(1/2)
Ε=178(3)^(1/2) /2
Ε=89(3)^(1/2)

[@Quest3669]

2x= √3-√2 & (1/2x)= √3+√2
(2x)^5 + (1/ 2x)^ 5= 178√3= K
?= K/2= 89√ 3 soln.

[@ManojkantSamal]

{(485 -198.*6)/4}+[(485+198.*6)/{(485)^2-(198.*6)^2}]........May be the half way of solution
Explain later...