Solve System: (x²+y²)/(x-y)² = 3 and (x-y)² = 1 for Product xy

To solve this problem, we'll follow these steps:

• Step 1: Substitute the given value of $(x-y)^2$ into the first equation.
• Step 2: Use the identity for a square of a difference to find a relationship between $x^2 + y^2$ and $xy$.
• Step 3: Solve for the product $xy$.

Now, let's work through each step:

Step 1: We start with the provided equation:

$\frac{x^2 + y^2}{(x-y)^2} = 3$

Given that $(x-y)^2 = 1$, we substitute:

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

which simplifies to:

$x^2 + y^2 = 3$

Step 2: We know from the identity of a square of a difference:

$(x-y)^2 = x^2 - 2xy + y^2$

Given $(x-y)^2 = 1$, we can write:

$x^2 - 2xy + y^2 = 1$

Step 3: We set up a system of equations:

$x^2 + y^2 = 3$ (Equation 1)

$x^2 - 2xy + y^2 = 1$ (Equation 2)

Subtract Equation 2 from Equation 1:

$(x^2 + y^2) - (x^2 - 2xy + y^2) = 3 - 1$

Simplifying the left side gives $2xy$:

$2xy = 2$

Divide both sides by 2:

$xy = 1$

Therefore, the product of $x$ and $y$ is $xy = 1$.