Solve system of equations $ y + 3x = 4x^3$ and $x + 3y = 4y^3$ Problem: In the set of real numbers find all $x,y$ that satisfy
$ y + 3x = 4x^3,x + 3y = 4y^3$.
What I don't understand, is how do I know that a condition satisfies both equations(I will show later exactly).
The solution goes like this: By addition and substraction of the equations we get the following 2 equations:
$x + y = (x + y)(x^2 − xy + y^2),x − y = 2(x − y)(x^2 + xy + y^2)$.
My question is, how do I know which conditions have to be true? According to the solution either $x+y=0$ or $x-y=0$, which I understand. My problem comes when they mention that both $(x^2 + xy + y^2)=1/2$ and $(x^2 - xy + y^2)=1$ have to be true at the same time. Why both? How do systems of equations work and how do we know that there won't be another solution if we multiply or divide the equations? Any tips for these kinds of problems are greatly appreciated.
Question 2: Why can't we also try to find a solution which satisfies for example $x-y=0$ and $(x^2 − xy + y^2)=1$
 A: Here is a natural way to solve the system. Continue with, 
$$x + y = (x + y)(x^2 − xy + y^2),\>\>\>\>\>
x − y = 2(x − y)(x^2 + xy + y^2)$$
or,
$$(x + y)(x^2 − xy + y^2-1)=(x − y)(x^2 + xy + y^2-\frac12)=0$$
So, there are four cases to be considered,
Case 1: Substitute $x-y=0$ into one of the original equations, say, $y + 3x = 4x^3$, which leads to,
$$4x(x^2-1)=0\implies x= 0, \pm 1$$
Case 2:  Similarly, $x+y=0$ leads to,
$$2x(2x^2-1)=0\implies x = 0,\pm \frac 1{\sqrt2}$$
Case 3: Substitute $x^2 − xy + y^2-1=0$ into $y+3x=4x^3$, leading to
$$16x^6-28x^4+13x^2-1=0,\>\> \text{or}$$
$$(x^2-1)(16x^4-12x^2+1)=0\implies x=\pm 1,\pm\frac{\sqrt5\pm 1}{4}$$
Case 4: Similarly $x^2 + xy + y^2-\frac12=0$ leads to
$$(2x^2-1)(16x^4-12x^2+1)=0\implies x=\pm \frac1{\sqrt2},\pm\frac{\sqrt5\pm 1}{4}$$
Thus, all the real solutions are:
$(0,0), (1,1), (-1,-1),
(\frac 1{\sqrt2},-\frac 1{\sqrt2}), (-\frac 1{\sqrt2},\frac 1{\sqrt2}), $
$(\frac {\sqrt5 + 1}{4},\frac {-\sqrt5 + 1}{4}), (\frac {\sqrt5 - 1}{4},\frac {-\sqrt5 - 1}{4}), 
(\frac {-\sqrt5 + 1}{4},\frac {\sqrt5 + 1}{4}), (\frac {-\sqrt5 - 1}{4},\frac {\sqrt5 - 1}{4})$
Note that the system of the two equations are implicitly of ninth order. So, it is expected to have nine pairs of solutions.
A: If $$a\cdot b =a\cdot b'$$ and $$c\cdot d=c\cdot d'$$ then if $a\neq 0 $ and $c\neq 0$ then $b=b'$ and $d=d'$.

For the second question, if $m\cdot n =0$ then we now only $m=0$ or $n=0$. It is not necessary both are $0$.
A: Hint.
We have with $f(x) = 4x^3-3 x$
$$
y = f(x)\\
x = f(y)\\
$$
or
$$
y = f^2(y)\\
x = f^2(x)
$$
so the solutions are the fixed points

which are
$$
\left\{0,\pm 1, \pm\frac{\sqrt 2}{2},\frac 14\left(-1\pm\sqrt 5\right),\frac 14\left(1\pm\sqrt 5\right)\right\}
$$
NOTE
$$
f^2(x)-x = f(f(x)) -x = 8 (x-1) x (x+1) \left(2 x^2-1\right) \left(4 x^2-2 x-1\right) \left(4 x^2+2 x-1\right)
$$
