\( 1 \leftarrow \) Find the product. \[ \frac{u^{3}+v^{3}}{u^{3}-v^{3}} \cdot \frac{u^{2}-v^{2}}{u^{2}+2 u v+v^{2}} \]

To find the product \[ \frac{u^{3}+v^{3}}{u^{3}-v^{3}} \cdot \frac{u^{2}-v^{2}}{u^{2}+2 u v+v^{2}}, \] we start by simplifying each factor. 1. The first factor, \(\frac{u^3 + v^3}{u^3 - v^3}\), can be simplified using the identities for the sums and differences of cubes: \[ u^3 + v^3 = (u + v)(u^2 - uv + v^2), \] \[ u^3 - v^3 = (u - v)(u^2 + uv + v^2). \] Therefore, we rewrite the first factor: \[ \frac{u^3 + v^3}{u^3 - v^3} = \frac{(u + v)(u^2 - uv + v^2)}{(u - v)(u^2 + uv + v^2)}. \] 2. The second factor, \(\frac{u^2 - v^2}{u^2 + 2uv + v^2}\), can be simplified as follows: \[ u^2 - v^2 = (u + v)(u - v), \] and \[ u^2 + 2uv + v^2 = (u + v)^2. \] Thus, we have: \[ \frac{u^2 - v^2}{u^2 + 2uv + v^2} = \frac{(u + v)(u - v)}{(u + v)^2} = \frac{u - v}{u + v}, \quad \text{for } u + v \neq 0. \] Now we can multiply the two simplified fractions together: \[ \frac{(u + v)(u^2 - uv + v^2)}{(u - v)(u^2 + uv + v^2)} \cdot \frac{(u + v)(u - v)}{(u + v)^2} = \frac{(u + v)(u^2 - uv + v^2)}{(u - v)(u^2 + uv + v^2)} \cdot \frac{(u - v)}{(u + v)}. \] Canceling \( (u + v) \) from numerator and denominator (since \( u + v \neq 0 \)): \[ = \frac{(u^2 - uv + v^2)}{(u - v)(u^2 + uv + v^2)}. \] Therefore, the final simplified product is: \[ \frac{u^2 - uv + v^2}{(u - v)(u^2 + uv + v^2)}. \]