China | Can you solve this ? | Math Olympiad | a=? & b=?

[@oahuhawaii2141]

a*√a + b*√b = 3 { E1 }
a*√b + b*√a = 2 { E2 }
Let x = √a and y = √b . Thus, a = x² and b = y² .
x³ + y³ = 3 { E1' }
x²*y + x*y² = 2 { E2' }
Note:
(x + y)³ = x³ + 3*x²*y + 3*x*y² + y³
= (x³ + y³) + 3*(x²*y + x*y²)
= 3 + 3*2 = 9 = 3²
x + y = ³√3² = s { E3 }
Use E3 in E2':
x*y*(x + y) = 2
x*y = 2/(x + y)
x*y = 2/s = p { E4 }
Use the quadratic formula with the sum & product:
z² - s*z + p = 0
z = (s ± √(s² - 4*p))/2
= (s ± √(s² - 4*2/s))/2
= (s ± √((s³ - 8)/s))/2
= (³√3² ± √(((³√3²)³ - 8)/³√3²))/2
= (³√3² ± √((3² - 8)/³√3²))/2
= (³√3² ± √(1/³√3²))/2
= (³√3² ± 1/³√3)/2
= (³√3³ ± 1)/2/³√3
= (3 ± 1)/2/³√3
= 4/2/³√3, 2/2/³√3
z = 2/³√3, 1/³√3
z² = 4/³√9, 1/³√9
= ³√3*4/3, ³√3/3
(x², y²) = (a, b) =
(³√3*4/3, ³√3/3), (³√3/3, ³√3*4/3)
He really should normalize the denominator to put it in standard form.

[@Zbigniew-b3u]

...można również wyjść z dodania i odejmowania obu równań stronami i bez żadnego podstawiania dojdziesz do tego samego wyniku.

[@Lupus_Cinereus]

👍

[@avalagum7957]

This is my solution before watching the video.
a^3 + b^3 = 3 = (a + b)((a + b)^2 - 3ab)
ab(a+b) = 2
s(s^2 - 3p) = 3
sp = 2
s^3 = 9

[@АндрейПергаев-з4н]

Когда получили
(х^2-ху+у^2)/(ху)=3/2
Просто можно поделить и тогда
х/у-1+у/х=3/2
Пусть х/у=с
с+1/с=5/2
Легко решить с= 2, с =1/2
у=2х или х=2у
Из уравнения 4
ху(х+у) =2
3х^3=1 или 3у^3=1
Дальнейшее понятно

[@oahuhawaii2141]

You should continue to solve for p, and then a and b of your version, and then find the a and b of the original set of equations.

[@bobraisbeck4185]

How do you know when to call it "exxie" and when just "ex"?

[@Lupus_Cinereus]

Nie masz wiekszych problemow? Moze w jego/jej jezyku tak sie x wymawia.

[@RealComplexity-math]

when its squared, its exxie