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.
 A: 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 $y$.
A: 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
A: Well, the second equation is a quadratic equation in $x$:
$$\text{a}\cdot x^2+\text{b}\cdot x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}{2\cdot\text{a}}\tag1$$
Now, in your example $\text{a}=\text{y},\text{b}=2\cdot\text{y}^2,\text{c}=\text{y}^3-2$, so we get:
$$x=\frac{-2\cdot\text{y}^2\pm\sqrt{\left(2\cdot\text{y}^2\right)^2-4\cdot\text{y}\cdot\left(\text{y}^3-2\right)}}{2\cdot\text{y}}=\pm\frac{\sqrt{2}}{\sqrt{\text{y}}}-\text{y}\tag2$$
