A Nice Maths Problem | X=? & Y=?

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How to solve for x and y in the systems of equations
x³ + y³ = 5 and x² + y² = 3
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[@GREAT.P]

❤

[@learncommunolizer]

❤️❤️❤️

[@juniorlopez1457]

I love your exercise ❤❤❤❤ I caught on you everything, but I've never thought about it😂😂😂

[@joseeoliviero6078]

I cannot solve this problem, however I'm positive this method described is the hardest path to solve it 😢

[@СергейКовалев-т1д6м]

Sometimes the hardest path is the only path...

[@debadityapurkyastha1777]

Fantastic

[@StephenRayWesley]

(3+2)= 5 (1+2)= 3

[@charlamps]

Beautifull

[@learncommunolizer]

Thank you very much! 😊👍

[@judhisthirbehera2972]

1-9+10=2 it is not -2 correct the mistake

[@bhattacharya86]

Very long sleeping it can be solved in four steps

[@oahuhawaii2141]

You made a bunch of minor errors. I'm surprised you gave up on finding all the solutions. I list all of them, using a more general approach.
*Background Info*
For z = x, y as roots of a quadratic equation:
(z - x)*(z - y) = 0
z² - (x + y)*z + x*y = 0 { Note sum and product }
z = ((x + y) ± √((x + y)² - 4*x*y))/2 { E1 }
= ((x + y) ± √(x² + 2*x*y + y² - 4*x*y))/2
= ((x + y) ± √(x² - 2*x*y + y²))/2
= ((x + y) ± √(x - y)²)/2
= ((x + y) ± (x - y))/2
z = x, y { For "±", the "+" yields x, the "-" yields y. }
Let s = (x + y) and p = (x*y) in E1:
z = (s ± √(s² - 4*p))/2 { E1' }
Let z⁺ = (s + √(s² - 4*p))/2
Let z⁻ = (s - √(s² - 4*p))/2
Thus, (x, y) = (z⁺, z⁻)
Since x & y are interchangeable, this means:
(x, y) = (z⁻, z⁺), (z⁺, z⁻) { E1" }
*Problem*
I list the squares as E2 and the cubes as E3:
x² + y² = 3 { E2 }
x³ + y³ = 5 { E3 }
Note: x & y are interchangeable.
Use E2 in the expansion of the square of the sum:
(x + y)² = x² + 2*x*y + y²
(x + y)² = (x² + y²) + 2*x*y
s² = 3 + 2*p
p = (s² - 3)/2 { E2' }
Use E3 in the expansion of the cube of the sum:
(x + y)³ = x³ + 3*x²*y + 3*x*y² + y³
(x + y)³ = (x³ + y³) + 3*x*y*(x + y)
s³ = 5 + 3*p*s { E3' }
Use E2' in E3' to eliminate p, and find s:
s³ = 5 + 3*((s² - 3)/2)*s
s³ = 5 + 3*s³/2 - 9*s/2
0 = 5 + s³/2 - 9*s/2
s³ - 9*s + 10 = 0
s³ - 4*s - 5*s + 10 = 0
s*(s - 2)*(s + 2) - 5*(s - 2) = 0
(s - 2)*(s² + 2*s - 5) = 0
s = 2, (-2 ± √24)/2
= 2, -1 ± √6
= 2, √6-1, -√6-1
Use E2' in E1', and substitute s:
z = (s ± √(s² - 4*(s² - 3)/2))/2
= (s ± √(s² - 2*(s² - 3)))/2
= (s ± √(6 - s²))/2
= (2 ± √2)/2, (√6-1 ± √(2*√6-1))/2,
(-(√6+1) ± i*√(2*√6+1))/2
From E1", we list the solutions:
(x, y) = (z⁻, z⁺), (z⁺, z⁻) =
(1 - √2/2, 1 + √2/2),
(1 + √2/2, 1 - √2/2),
((√6-1 - √(2*√6-1))/2, (√6-1 + √(2*√6-1))/2),
((√6-1 + √(2*√6-1))/2, (√6-1 - √(2*√6-1))/2),
(-(√6+1)/2 - i*√(2*√6+1)/2, -(√6+1)/2 + i*√(2*√6+1)/2),
(-(√6+1)/2 + i*√(2*√6+1)/2, -(√6+1)/2 - i*√(2*√6+1)/2)

[@souzasilva5471]

Faltou destacar as respostas, por esse motivo, não gostei.

[@josuefigueroamendoza3705]

Ojo: Utiliza la misma variable, para diferentes funciones.confunde al alumno

[@nagarajahshiremagalore226]

Please don't make mistake while writing.

[@windyyw]

u dont finish it.

[@oahuhawaii2141]

I found the 6 solutions for you.