135. \( \left(\frac{x^{2} y^{-2}}{x^{-3} y}\right)^{2}=x^{10} y^{2} \)

To solve the equation \( \left(\frac{x^{2} y^{-2}}{x^{-3} y}\right)^{2}=x^{10} y^{2} \), we start by simplifying the left side. 1. Simplify the fraction inside the parentheses: \[ \frac{x^{2} y^{-2}}{x^{-3} y} = x^{2 - (-3)} y^{-2 - 1} = x^{2 + 3} y^{-2 - 1} = x^{5} y^{-3} \] 2. Now we can raise this simplified expression to the power of 2: \[ (x^{5} y^{-3})^2 = x^{5 \cdot 2} y^{-3 \cdot 2} = x^{10} y^{-6} \] 3. Now, our equation looks like this: \[ x^{10} y^{-6} = x^{10} y^{2} \] 4. To solve for \(y\), we can set the \( y \) components equal to one another (since the \(x^{10}\) terms are the same on both sides): \[ y^{-6} = y^{2} \] 5. Rewrite this equation as: \[ \frac{1}{y^{6}} = y^{2} \] Which can be rearranged to: \[ 1 = y^{2} \cdot y^{6} = y^{8} \] 6. Hence, \(y^{8} = 1\). The real solutions to this equation would be: \[ y = 1 \quad \text{or} \quad y = -1 \] Thus, the solutions to the original equation are \(y = 1\) or \(y = -1\). If we require \(x\) to be any real number, the final answers become: \[ \text{For } y = 1: \quad x \text{ can be any real number.} \] \[ \text{For } y = -1: \quad x \text{ can also be any real number.} \] So the complete solution set is \( (x, y) \) where \( x \) is any real number and \( y \) is either 1 or -1.