x^3y^3 - y = x

To solve for y in terms of x in the equation x^3y^3 - y = x, we can first factor out y from the left side of the equation:

y(x^3y^2 - 1) = x

Next, divide both sides by (x^3y^2 - 1) to isolate y:

y = x / (x^3y^2 - 1)

Therefore, y = x / (x^3y^2 - 1) is the solution for y in terms of x.

$$x^3y^3 - y = x$$

$$x^3y^3 - y - x = 0$$

$$x^3y^3 - y^3 - x + y^3 = 0$$

$$y^3(x^3 - 1) - (x - y^3) = 0$$

$$y^3(x - 1)(x^2 + x + 1) - (x - y^3) = 0$$

$$(x - 1)(y^3(x^2+x+1) - 1) = 0$$

$$(x - 1)(y^3x^2 + y^3x + y^3 - 1) = 0$$

Either $$x - 1 = 0$$ or $$y^3x^2 + y^3x + y^3 - 1 = 0$$

Case 1:

$$x - 1 = 0$$

$$x = 1$$

Case 2:

$$y^3x^2 + y^3x + y^3 - 1 = 0$$

Method 1: Rational Root Theorem

Possible rational roots: ±1

Trying \(y = 1\):

$$1^3x^2 + 1^3x + 1^3 - 1 = 0$$

$$x^2 + x = 0$$

$$x(x + 1) = 0$$

$$x = 0, -1$$

Trying \(y = -1\):

$$(-1)^3x^2 + (-1)^3x + (-1)^3 - 1 = 0$$

$$x^2 - x = 0$$

$$x(x - 1) = 0$$

$$x = 0, 1$$

So, the rational roots of the equation are \(x = 0, -1, 1\).

Method 2: Factoring

$$y^3x^2 + y^3x + y^3 - 1 = 0$$

$$y^3(x^2 + x + 1) - 1 = 0$$

$$(y^3 - 1)(x^2 + x + 1) = 0$$

$$(y - 1)(y^2 + y + 1)(x^2 + x + 1) = 0$$

Since \(y^2 + y + 1\) and \(x^2 + x + 1\) are both irreducible quadratic expressions, the solutions are:

$$y = 1, \quad x = \text{any real number}$$

$$y = -1, \quad x = \text{any real number}$$

So, the solutions to the equation \(x^3y^3 - y = x\) are:

$$\boxed{x = 1, \quad y = \text{any real number}}$$

$$\boxed{x = -1, \quad y = \text{any real number}}$$

$$\boxed{x = 0, \quad y = 1}$$

$$\boxed{x = 0, \quad y = -1}$$