Solve sqrt{x^1+y^3}=2

$$\sqrt{ { x }^{ 1 } + { y }^{ 3 } } =2$$

Solve for x

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

$$y\geq -\sqrt[3]{4-y^{3}}$$

Steps for Solving Radical Equations

Calculate x to the power of 1 and get x.

$$\sqrt{x+y^{3}}=2$$

Square both sides of the equation.

$$x+y^{3}=4$$

Subtract y^{3} from both sides of the equation.

$$x+y^{3}-y^{3}=4-y^{3}$$

Subtracting y^{3} from itself leaves 0.

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

Solve for x (complex solution)

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

Solve for y (complex solution)

$$y=e^{\frac{2i\pi }{3}}\sqrt[3]{4-x}$$

$$y=\sqrt[3]{4-x}$$

$$y=e^{\frac{4i\pi }{3}}\sqrt[3]{4-x}$$

Solve for y

$$y=\sqrt[3]{4-x}$$