An Interesting Algebra Challenge | Give It A Try!

[@RashmiRay-c1y]

After some manipulations, x=3[1/(√7 + 1) + 1/(√13 +√7)] = 1/2(√13 -1). Thus, x^2=3-x, x^4=12-7x, x^6=57-40x. Therefore, x^6+40x-57=0. S0, k=57.

[@gregevgeni1864]

K = 57
Because 1/x = (1+√13)/6 =>
x = (√13-1)/2 => x⁶ =(x³)²=
(2√13-5)²=77-20√13 .
Now
x⁶+40x-k=0 =>
(77-20√13)+40(√13-1)/2-k=0 =>
k= 77-20√13+20√13-20=57.

[@Fjfurufjdfjd]

κ=57

[@SKASKumar]

K=57