Thailand Math Olympiad | A Nice Algebra Challenge 

After you derive x=(-1+sqrt(13))/6 I think it is much faster to just multiply it out and show x^2=x*x=(7-sqrt(13))/18, x^3=x*x^2=(-5+2*sqrt(13))/27, x^6=x^3*x^3=(77-20*sqrt(13))/27^2

Respected Sir, Pranam. The way, which has been taken by you to reach at the decisive end of the solution is exemplary and admirable

Beautiful

Beautiful problem. I loved and enjoyed doing it.

Threw me for a sec. It looked like 272 but you meant 27^2.

X= (root 13 -1)/6, but unable to find out the answer of the given question. ^=read as to the power *= read as square root As per question numerator is 7+*91+7.*7+7.*13 =(7+*7.*13)+(7.*7+7.*13) =*7(*7+*13)+7(*7+*13) =(*7+*13)(*7+7) Let a=*7+*13, b=*7+7 So numerator is 'a+b' Now explain the denominator 14+*91+*7 =7+7+*91+*7 =(7+*91)+(7+*7) =*7(*7+*13)+(7+*7) =*7a+b As per question 1/x=ab/(*7a+b) So, x=(*7a+b)/ab =(*7a/ab)+(b/ab) =(*7/b)+(1/a) Let's explain "*7/b" *7/(*7+7)=*7/(7+*7) ={*7(7-*7)}/{(7+*7)(7-*7)} (7.*7-7)/{7^2-*7^2} =(7.*7-7)/(49-7) =(7.*7-7)/42 Now explain 1/a=1/(*13+*7) =1(*13-*7)/{(*13+*7)(*13-*7) =(*13-*7)/(*13^2-*7^2) =(*13-*7)/(13-7)=(*13-*7)/6 Now, (*7/b)+(1/a) ={(7.*7-7)/42}+{(*13-*7)/6} ={7.*7-7+7.*13-7.*7}/42 =(7.*13-7)/42 =7(*13-1)/42 =(*13-1)/6

Let sqrt(7)=a and sqrt(13)=b. Then, x=[a+1+a+b]/(a+1)(a+b) = 1/(a+b+ + 1/(a+1) = (sqrt(13)-1)/6. So, x^6= [4928 -1280sqrt(13)]/6^6 = [77-20sqrt(13)]/729.

X⁶=(77-20√13)/729

X^6= (77- 20√13)/729

Marathon problem! Nice choice and nicely solved - even though the final result isn't so pretty! Thank you.

(x ➖ 1x+1) 1 +2^31+1 1 13^1/2^7 +3^31+1 1^1^1 1^1 /1^1 1^1^1 1^1 (x ➖ 1x+1) x^3^2 (x ➖ 3x+2)

(x ➖ 1x+1) 1 +3^31+ 1 1+ 1 13^1/2^7 3^31 1 1^1^1 1^1/2^1 1^1^1 1^1/2^1 2^1 (x ➖ 2x+1 ) x^3^2 (x ➖ 3x+2)

10:20 Четыре слагаемых, в следующей записи три Куда то исчезло 3х^2, оно было последнее , и его не заменили на 1-х, а просто убрали неизвестно как

[77-20(ριζα13)]/729