Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$ 
Find the real solutions for the system:
  $$\left\{
\begin{array}{l}
x^3+y^3=1\\ 
x^2y+2xy^2+y^3=2\\ 
\end{array}
\right. 
$$

From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqrt[3]{2}})$ and $(\dfrac{1}{3^{\frac{2}{3}}},\dfrac{2}{3^{\frac{2}{3}}})$.
Working with the expressions I could find that an equivalent system is 
$$\left\{
\begin{array}{l}
(x+y)(x^2-xy+y^2)=1\\ 
y(x+y)^2=2\\ 
\end{array}
\right. 
$$
Developing these expressions I got stuck. 
Hints and answers are appreciated. Sorry if this is a duplicate.

Hint:
Observe that $xy\ne0$
Set $y=mx$
divide the resultant one equation by the other to form a cubic equation in $m$ with $m+1\ne0$ being one factor
