Equation with an infinite number of solutions I have the following equation: $x^3+y^3=6xy$. I have two questions: 1. Does it have an infinite number of rational solutions? 
2. Which are the solutions over the integers?($ x=3 $ and $ y=3 $ is one)
Thank you!
 A: If $x$ and $y$ are rational, then so is $y/x=\alpha$. Then $x^3+y^3=6xy$ becomes
$$
(\alpha^3+1)x^3-6\alpha x^2=0\tag{1}
$$
and then $x=0$ or $x=\dfrac{6\alpha}{\alpha^3+1}$. Thus, for any rational $\alpha$, we have the rational solutions
$$
\left(\frac{6\alpha}{\alpha^3+1},\frac{6\alpha^2}{\alpha^3+1}\right)\tag{2}
$$

Since $x+y=\dfrac{6\alpha}{\alpha^2-\alpha+1}$, we have that
$$
-2\lt x+y\le6\tag{3}
$$
Note that $x+y=-2$ only happens when $\alpha=-1$, and that doesn't give a finite $(x,y)$ in $(2)$.
Thus, cubing $(3)$ yields
$$
-8\le x^3+3x^2y+3xy^2+y^3\le216\tag{4}
$$
and applying $x^3+y^3=6xy$,
$$
-\frac83\le(x+y+2)xy\le72\tag{5}
$$
Since $x+y+2\ge1$, inequality $(5)$ leaves only a finite number of $x,y\in\mathbb{Z}$ to check.
A: 
Which are the solutions over the integers?

$(x+y)^3=\underbrace{x^3+y^3}_{6xy}+3xy(x+y)=3xy(x+y+2)\iff6|(x+y)$, since, on one hand, $3$ is a prime, and, on the other hand, x and y being of opposite parity would lead to contradiction. Hence, we have $x=2a+r$, and $y=2b-r$, with $r\in\{0,1\}$, and, at the same time, $x=3A+R$, and $y$$=3B-R$, also with $R\in\{0,1\}$. Can you take it from here ? :-)
A: Wolfram Alpha says that there are no rational solutions except the one you noted, $x=y=3$ although.
It seems that it chose to skip the trivial $x=y=0$ though. The link has some irrational solutions too, if you need them.
A: The equation is symmetric in $x$ and $y,$ suggesting solutions of the form $x=y.$ Plugging that in, we find $$2x^3-6x^2=0$$ This gives the solution $x=y=0.$ 
Simplifying by dividing by $2x^2$ gives $x-3=0,$ leading to another solution $x=y=3.$ 
