Question

determine x+y onde x e y são reais, sabendo que x³+y³=2 e x²y+xy²=5.

Answer 1

$\left(x+y \right )^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}\\ \\ \left(x+y \right )^{3}=\underbrace{x^{3}+y^{3}}_{2}+3\cdot \left( \right.\underbrace{x^{2}y+xy^{2}}_{5}\left.\right )\\ \\ \left(x+y \right )^{3}=2+3\cdot 5\\ \\ \left(x+y \right )^{3}=2+15\\ \\ \left(x+y \right )^{3}=17\\ \\ x+y=\sqrt[3]{17}$

Answer 2

  x³+y³= 2    x²y+xy² = 5.
( x + y)³ = x³ + 3x²y + 3xy² + y³

x³ + y³ + 3(x²y + xy² ) ==>  2 + 3.5  ==> 17

  (x + y)³ =  17   ==> ( x  + y ) = ∛17