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[@Quest3669]

X^1/3+(1/x^1/3)= 1298
let ? = t ; cubing t gives
t^3- 3t- 1298= 0 or
? = t = 11 for real soln.

[@潘博宇-k4l]

E=11.

[@RashmiRay-c1y]

Let x^1/18=t. So, t^3-1/t^3=36. E = t^2+1/t^2. Let a=t-1/t. So, a^3= 36-3a > a=3 is the only real solution. Thus, t^2+1/t^2-2=9 > E=11.

[@yakupbuyankara5903]

11

[@vremiavremiavremiavremia1706]

😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊😊

[@ManojkantSamal]

^=read as to the power
*=read as square root
As per question
X^(1/6) - {1/x^(1/6)}
X^(1/6)={x^(1/18)}^3
Let x^(1/18)= r
So,
r^3 - (1/r)^3=36
{r-(1/r)}^3 + {3×r×(1/r)}{r-(1/r)}=36
Let r-(1/r)=p
So,
P^3+3p=36
Through H &T method I got that
P=3
So,
r-(1/r)=3
Take the square
r^2+(1/r)^2 +{2×r×(1/r)}=3^2
r^2+(1/r)^2+2=9
r^2+(1/r)^2=9-2=7........eqn1
Now explain
r^2={x^(1/18)}^2=x^(1/9)
X^(1/9)+{1/x^(1/9)}=7

[@RealQinnMalloryu4]

(x)^6 ➖( 1)^2/(x)^6=x^6 ➖ 1/x^6 x^5/x^6 =x^1(x ➖ 1x+1).{x+x+x+x+x+x+x+x+xx ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x}=x^9+{1+1 ➖ }/{x+x+x+x+x+x+x+x+xx ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x ➖ x}={x^9+2}/x^9=2x^9/x^9=2x^1 (x ➖ 2x+1).

[@Fjfurufjdfjd]

Θετω (χ)^(1/6)=α>0 χ>0 Ε>0. Εχω (χ^2)^(1/6)=α^2 ή [(χ)^(1/9)]^3=α^2. Αρα Ε=(χ)^(1/9)+1/[(χ)^(1/9)]. Απο τη σχεση που δινεται εχω α-(1/α)=36 ; α^2+[(1/α)^2]=36^2+2=1298. Ε^3-3Ε-1298=0 ; (Ε-11)(Ε^2+11Ε+118)=0.[Δ

[@vremiavremiavremiavremia1706]

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[@giuseppemalaguti435]

11