Solve xy=frac{sqrt{x^2-1}}{sqrt[3]{x^3-1}}

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

Solve for y

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

$$x\leq -1\text{ or }x>1$$

Steps for Solving Linear Equation

The equation is in standard form.

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

Divide both sides by x.

$$\frac{xy}{x}=\frac{\sqrt{x^{2}-1}}{\sqrt[3]{\left(x-1\right)\left(x^{2}+x+1\right)}x}$$

Dividing by x undoes the multiplication by x.

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

Divide \frac{\sqrt{x^{2}-1}}{\sqrt[3]{\left(x-1\right)\left(x^{2}+x+1\right)}} by x.

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