Making the equation $x^3+1=3xy-y^3$ explicit

Question

$$x^3+1=3xy-y^3$$
How do I turn this equation into the form of $y=f(x), x\neq1$? When I graph the above, it looks like $y=-x-1$.
I've tried factoring and can't seem to make it work.

Note: The original question asked for the closest distance between $x^3+1=3xy-y^3$ and $(6,7)$. I could probably substitute values into the equation to see how it looks like and realise that it looks like a line, but I want to try to get an explicit equation.

Answer 1

There exist the following formula of Euler:

If $$ x^3+y^3+z^3=3xyz\Leftrightarrow (x+y+z=0\textrm{ or }x=y=z) $$ Hence you equation is written $$ x^3+y^3+1^3=3x y\cdot 1\Leftrightarrow (x+y+1=0\textrm{ or }x=y=1) $$ QED

REVISED. The formula of Euler read as $$ x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2] $$ If one makes the evaluations it easily follows.

Answer 2

Let $y:=u-x-1$. Then

$$y^3-3xy+x^3+1=u(u^2-3(x+1)u+3x^2+3x+3)$$

The root $u=0$ is of course $$y=-x-1.$$ Then the discriminant of the quadratic equation is

$$-3(x-1)^2$$ and the only real solution is

$$x=1\implies y=1.$$