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[@RashmiRay-c1y]

Note that 1/x=√11 +√5. So, 3x+1/(2x)=√11. Let 3x=a and 1/(2x)=b. Then, a+b=√11 and ab=3/2 and E=(a+b)^5-5ab(a^3+b^3)-10 a^2b^2(a+b). But a^3+b^3=(a+b)[(a+b)2-3ab] = (13/2)√11. So, E = √11[121-195/4 -45/2] = (199/4)√11.

[@Quest3669]

3x+(1/2x)= √11 hence
? = (199√11)/ 4= 165 approx.

[@RealQinnMalloryu4]

(11)^2 ➖ (5)^2/(6)={121 ➖ 25}/36=06/362.24 2.2^12 1.2^61.2^2^3 1.1^2^3 (x ➖ 3x+2). {243x^5+243x^5 ➖}{1x+1x ➖ }/{32x^5+32x^5 ➖ }={486x^102x^2}/64x^10=488x^12/64x^10=802x^2 10^82 10^2^42x^2.10^2^21x^2 10^2^3^7x^2 2^5^1^3^3^4x^2 1^1^1^3^2^2x^2 3^1^1x^2 3x^2 (x ➖ 3x+2).

[@gregevgeni1864]

K = (199√11)/4
x = (√11-√5)/6 = 1/(√11+√5) =>
(3x) + 1/(2x) = 3/(√11+√5) +
(√11+√5)/2=(√11-√5)/2+(√11+√5)/2
=> (3x)+1/(2x)= √11.
Now
(3x)³+(1/2x)³=(3x+1/2x)³-3•3x•1/2x•(3x+1/2x) = (√11)³-(9/2)•√11= 13√11/2 (1)
(3x)²+(1/2x)²= (3x+1/2x)²-2•3x•1/2x=
= (√11)²-3=8 (2).
(1)•(2) =>
{(3x)³+(1/2x)³}•{(3x)²+(1/2x)²}= 4•13√11 =>
243x⁵ + 1/(32x⁵) + (9/4)•(3x+1/2x)=52√11 =>
243x⁵+1/(32x⁵)= 52√11-(9/4)√11=
K = 199√11/4

[@adribber]

x = (√11 - √5)/6
243x⁵ + 1/32x⁵ = ?
6x = √11 - √5
Let, a = 3x , b = 2x
a = (√11 - √5)/2 , b = (√11 - √5)/3
a + 1/b = (√11 - √5)/2 + 1/(√11 - √5)/3
= (√11 - √5)/2 + 3/(√11 - √5)
= (√11 - √5)/2 + 3(√11 + √5)/(√11 - √5)(√11 + √5)
= 3(√11 - √5)/6 + 3(√11 + √5)/6
= √11
a/b = (√11 - √5)/2 / (√11 - √5)/3
= 3(√11 - √5)/2(√11 - √5)
= 3/2
Let, k = 243x⁵ + 1/32x⁵
k = (3x)⁵ + 1/(2x)⁵
= a⁵ + 1/b⁵
= (a + 1/b)(a⁴ - a³/b + a²/b² - a/b³ + 1/b⁴)
= (a + 1/b)(a⁴ + 1/b⁴ - a³/b - a/b³ + a²/b²)
= (a + 1/b){a⁴ + 1/b⁴ - a/b(a² + 1/b²) + a²/b²}
= (a + 1/b)[{(a² + 1/b²)² - 2a²/b²} - a/b{(a + 1/b)² - 2a/b} + a²/b²}]
= (a + 1/b)[{(a + 1/b)² - 2a/b}² - a/b(a + 1/b)² + a²/b²]
= (a + 1/b)[(a + 1/b)⁴ - 5a/b(a + 1/b)² + 5a²/b²]
= √11(√11⁴ - 5*(3/2)√11² + 5*9/4)
= √11(484 - 330 + 45)/4
= 199√11/4