An Engaging Algebra Problem | Can You Crack This Challenge?

[@pietergeerkens6324]

Nice problem!
I used Simon's Favourite Factoring Trick, multiplying through by 8 the cubic obtained by clearing denominators and isolating x, to obtain:
(2x)³ = 8*x³
= 8*7√53 - 8*20
= 56√53 - 160
= 53√53 + 3*√53 - 3*53 - 1
= 53√53 - 3*53 + 3*√53 - 1
= (√53 - 1)³
and
x = ½ * (√53 - 1).
Sometimes simpler is better.

[@Shobhamaths]

X=(√53-1) /2
x^3=7√53-20
y^3=7√53+20
x^3-y^3=-40, by formula (x-y) (x^2+xy+y^2) =-40
t=x-y
t^3+39t+40=0
t=-1
x-y=-1;
x+y=±√53;solve them
x=(√53-1) /2
xy=13

[@Fjfurufjdfjd]

χ=[(53)^(1/2)-1]/2

[@joelstonestreeh9263]

Where does the Y come from?

[@woobjun2582]

taking the conjugate of x^3