<s>[INST] 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.
 [/INST] Multiply the first equation by $2$ and then set the two left sides equal:
$$2x^3+2y^3 = x^2y+2xy^2+y^3.$$
This is a homogeneous equation of degree $3$, so divide through by $x^3$:
$$2+2\frac{y^3}{x^3} = \frac{y}{x}+2\frac{y^2}{x^2}+\frac{y^3}{x^3}.$$
Substitute $u=y/x$:
$$2+2u^2=u+2u^2+u^3$$
$$u^3-2u^2-u+2=0$$
$$(u^2-1)(u-2)=0$$
$$u=-1, 1 \mbox{ or } 2.$$
Dividing the first equation by $x^3$ gives $1+u^3 = 1/x^3.$  Plug each value of $u$ into this to find values of $x$ and then $y$.
</s>