An Interesting Olympiad Challenge | Can You Solve This?

[@ManojkantSamal]

Respected Sir, Good evening

[@RashmiRay-c1y]

We are given that (x^6+1)/x^3 = 52. So, x is positive and x^3+1/x^3=52. Now x^3+1/x^3 = (x^3/2 + x^-3/2)^2 -2 > x^3/2 + x^-3/2 = [52 + 2}^1/2 = 3√6. So, [x^3/2 + x^-3/2 ]^3 = x^9/2 + x^-9/2 + 3 (x^3/2 + x^-3/2 ) > x^9/2 + x^-9/2 = (3√6)^3 - 9√6 = 153√6. We have to evaluate E = x^9/2/(1+x^9) = [x^9/2 + x^-9/2]^-1 = 1/153√6. So, E = 1/(153√6).

[@kassuskassus6263]

x=1/153sqrt6 or x= sqrt6/918

[@Quest3669]

?= √1/(140454)= 1/(153√6)

[@satrajitghosh8162]

Writing z for x^ (3/2) one gets
z ^2 + 1/z^2 = 52
(z + 1/z)^2 - 2 = 52
z + 1/z = √ (54)
Hereby
z^3 + 1/z^3
= ( z + 1/z)^3 - 3 z * (1/z)( z + 1/z)
= 54 ^ (3/2) - 3 √ (54)
= 51 * √ (54)
= 153 * √6
Hereby
z^3 /( 1 + z^6) = 1/( 153 √6)

[@alexnikola7520]

x^9=t^6... t^2/(1+t^4)=1/52...t^2+1/t^2=52... t^3/(1+t^6)=?...t^3+1/t^3=1/?=(t+1/t)(t^2+1/t^2-1)=(t+1/t)*51... (t+1/t)^2-2=52...t+1/t=√54=3√6...?=1/51*3√6

[@xualain3129]

x^3/(1+x^6)=1/52 ->(1+x^6)/x^3=52 ->x^3+1/x^3=52
Let sqrt(x^9)/(1+x^9)=t ->x^9/(1+2*x^9+x^18)=t^2 ->x^9+1/x^9+2=1/t^2
(x^3+1/x^3)^3-3*x^3*(1/x^3)*(x^3+1/x^3)=1/t^2
52^3-3*52=1/t^2 -->t^2=1/140454 t=1/(153*sqrt(6)) discarding the negative for t is positive because x^3 is positive and so is 1+x^9
x^3=(52+sqrt(52^2-4))/2 or (52-sqrt(52^2-4))/2 both of which are positive.

[@RealQinnMalloryu4]

{x^3+x^3 ➖ }/{1+1 ➖}{x^6+x^6 ➖}=x^6/{2+x^12}=x^6/2x^12=2x^6 2x^3^2 1x1^2 x^1^2 (x ➖ 2x+1).{x^9+x^9 ➖ }/{1+1 ➖ }+{x^9+x^9 ➖ }=x^18/{2+x^18}=x^18/2x^18=2x^1 (x ➖ 2x+1).

[@gaiatetuya92]

問題に対してもっと簡素な解答になるようにしろよ。