Solving A System of Equations 

What was most interesting to me was that even though you had cubic terms, this system is really a quadratic polynomial in disguise. The x^3 terms cancel and there are only two (complex) solutions. Not three solutions as you might suspect.

x + y = 4 x³ + y³ = 4 But x³ + y³ = (x + y)(x² - xy + y²) 4 = 4(x² - xy + y²) x² + y² - xy = 1 x² + y² + 2xy = 16 3xy = 15 xy = 5 x = 5/y y + 5/y = 4 y² + 5 = 4y y² - 4y + 5 = 0 y = 2 ± √(4 - 5) y = 2 ± i As the equation set is symmetric about the variables, x is also 2 ± i, but inverted to y. Therefore, (x,y) = {(2 + i, 2 - i), (2 - i, 2 + i)}

Happy new year, friend. Feliz ano novo

At first glance, you don't expect to see such a result 😊 Happy New Year, dear math lovers!!

Happy new Years 🎉

Working even on holidays !

(x+y)^3=x^3+y^3-3xy 4^3=4-3*4xy Xy=5 x+y=4 Now solve u^2-4u+5=0 and the roots are interchangeably x and y

My steps: 1) (x+y)^3 = 64, subtract x^3 + y^3 2) 3(x^2*y+x*y^2) = 60 3) x*y(x+y) = 20, (x+y) = 4 gives 4) x*y = 5. since x =4-y, we get 5) y^2 -4y +5 = 0. This solves for {2-i,2+i}. Thanks!

I was so close - I just forgot to add 2xy to both sides after completing the square.

This polynomial system looks suspiciously familiar!

This can be solved with a simple complex substitution that translates the polys in x and y to make them nice in C. x + y = x^3 + y^3 (= 4) => -x^3 + x = y^3 - y => -x (x^2 - 1) = y(y^2 - 1) => -(x - 1)x(x + 1) = (y - 1)y(y + 1) x + y = 4 => y = 4 - x Let z = y - 2 = 4 - x - 2 = 2 - x => x = 2 - z, y = 2 + z => -(2 - z - 1)(2 - z)(2 - z + 1) = (2 + z - 1)(2 + z)(2 + z + 1) => -(1 - z)(2 - z)(3 - z) = (1 + z)(2 + z)(3 + z) => (z - 1)(6 - 2z - 3z + z^2) = (z + 1)(6 + 2z + 3z + z^2) => (z - 1)(z^2 -5z + 6) = (z + 1)(z^2 + 5z + 6) => z^3 - 5z^2 + 6z - z^2 + 5z - 6 = z^3 + 5z^2 + 6z + z^2 + 5z + 6 => z^3 - 6z^2 + 11z - 6 = z^3 + 6z^2 + 11z + 6 => 12z^2 + 12 = 0 => z^2 + 1 = 0 => z = +/- i => x = 2 -+ i, y = 2 +- i I like this problem. It's nice and concise and has a straightforward solution.

(2+i; 2-i); (2-i; 2+i)

x = 2 + i, y = 2 - i x = 2 - i, y = 2 + i not real numbers (imaginary)