Solve (x+2)(y^2-2y+4)=y^3+2y

$$( x + 2 ) ( y ^ { 2 } - 2 y + 4 ) = y ^ { 3 } + 2 y$$

Solve for x

$$x=-\frac{8-6y+2y^{2}-y^{3}}{y^{2}-2y+4}$$

Steps for Solving Linear Equation

Use the distributive property to multiply x+2 by y^{2}-2y+4.

$$xy^{2}-2xy+4x+2y^{2}-4y+8=y^{3}+2y$$

Subtract 2y^{2} from both sides.

$$xy^{2}-2xy+4x-4y+8=y^{3}+2y-2y^{2}$$

Add 4y to both sides.

$$xy^{2}-2xy+4x+8=y^{3}+2y-2y^{2}+4y$$

Combine 2y and 4y to get 6y.

$$xy^{2}-2xy+4x+8=y^{3}+6y-2y^{2}$$

Subtract 8 from both sides.

$$xy^{2}-2xy+4x=y^{3}+6y-2y^{2}-8$$

Combine all terms containing x.

$$\left(y^{2}-2y+4\right)x=y^{3}+6y-2y^{2}-8$$

The equation is in standard form.

$$\left(y^{2}-2y+4\right)x=y^{3}-2y^{2}+6y-8$$

Divide both sides by y^{2}-2y+4.

$$\frac{\left(y^{2}-2y+4\right)x}{y^{2}-2y+4}=\frac{y^{3}-2y^{2}+6y-8}{y^{2}-2y+4}$$

Dividing by y^{2}-2y+4 undoes the multiplication by y^{2}-2y+4.

$$x=\frac{y^{3}-2y^{2}+6y-8}{y^{2}-2y+4}$$