Solve x^3+3y-xxy/4=0

$$x ^ { 3 } + 3 y - x \frac { x y } { 4 } = 0$$

Solve for y

$$y=-\frac{4x^{3}}{12-x^{2}}$$

$$|x|\neq 2\sqrt{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$4\left(x^{3}+3y\right)-xxy=0$$

Use the distributive property to multiply 4 by x^{3}+3y.

$$4x^{3}+12y-xxy=0$$

Multiply x and x to get x^{2}.

$$4x^{3}+12y-x^{2}y=0$$

Subtract 4x^{3} from both sides. Anything subtracted from zero gives its negation.

$$12y-x^{2}y=-4x^{3}$$

Combine all terms containing y.

$$\left(12-x^{2}\right)y=-4x^{3}$$

Divide both sides by 12-x^{2}.

$$\frac{\left(12-x^{2}\right)y}{12-x^{2}}=-\frac{4x^{3}}{12-x^{2}}$$

Dividing by 12-x^{2} undoes the multiplication by 12-x^{2}.

$$y=-\frac{4x^{3}}{12-x^{2}}$$