Solve x^2+(ysqrt[3]{x^2})=1

$$x ^ { 2 } + ( y \sqrt[ 3 ] { x ^ { 2 } } ) = 1$$

Solve for y

$$y=-x^{\frac{4}{3}}+\frac{1}{x^{\frac{2}{3}}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$y\sqrt[3]{x^{2}}=1-x^{2}$$

The equation is in standard form.

$$\sqrt[3]{x^{2}}y=1-x^{2}$$

Divide both sides by \sqrt[3]{x^{2}}.

$$\frac{\sqrt[3]{x^{2}}y}{\sqrt[3]{x^{2}}}=\frac{1-x^{2}}{\sqrt[3]{x^{2}}}$$

Dividing by \sqrt[3]{x^{2}} undoes the multiplication by \sqrt[3]{x^{2}}.

$$y=\frac{1-x^{2}}{\sqrt[3]{x^{2}}}$$

Divide -x^{2}+1 by \sqrt[3]{x^{2}}.

$$y=-x^{\frac{4}{3}}+\frac{1}{x^{\frac{2}{3}}}$$