Solve y^2+z^3-x=1

$$y ^ { 2 } + z ^ { 3 } - x = 1$$

Solve for x

$$x=y^{2}+z^{3}-1$$

Steps for Solving Linear Equation

Subtract y^{2} from both sides.

$$z^{3}-x=1-y^{2}$$

Subtract z^{3} from both sides.

$$-x=1-y^{2}-z^{3}$$

The equation is in standard form.

$$-x=1-z^{3}-y^{2}$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{1-z^{3}-y^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{1-z^{3}-y^{2}}{-1}$$

Divide -y^{2}-z^{3}+1 by -1.

$$x=y^{2}+z^{3}-1$$

Solve for y

$$y=\sqrt{x-z^{3}+1}$$

$$y=-\sqrt{x-z^{3}+1}\text{, }z\leq \sqrt[3]{x+1}$$