1. **Problem statement:** We have two cylindrical rods with radii $x$ cm and $y$ cm, held tightly together by a thin band of length $P$ cm. We need to show that $$P = 4\sqrt{xy} + \pi(x + y) + 2(x - y)\sin^{-1}\left(\frac{x - y}{x + y}\right).$$ 2. **Understanding the problem:** The band wraps around the two circles, touching them externally. The length $P$ consists of two straight segments and two arcs from the circles. 3. **Key geometry:** The distance between the centers of the two circles is $x + y$ because they touch externally. 4. **Length of straight segments:** The band touches the circles at points forming two common external tangents. The length of each straight segment is $2\sqrt{xy}$, so total straight length is $4\sqrt{xy}$. 5. **Length of arcs:** The band also covers arcs on each circle. The angle subtended by the arcs is $2\theta$, where $$\theta = \sin^{-1}\left(\frac{x - y}{x + y}\right).$$ 6. **Arc lengths:** The arc length on the circle with radius $x$ is $(\pi - 2\theta)x$ and on the circle with radius $y$ is $(\pi - 2\theta)y$. 7. **Total arc length:** Adding these, $$ (\pi - 2\theta)(x + y) = \pi(x + y) - 2\theta(x + y).$$ 8. **Combine all parts:** Total length $P$ is sum of straight segments and arcs: $$P = 4\sqrt{xy} + \pi(x + y) - 2\theta(x + y).$$ 9. **Rewrite the last term:** Since $\theta = \sin^{-1}\left(\frac{x - y}{x + y}\right)$, $$-2\theta(x + y) = 2(x - y)\sin^{-1}\left(\frac{x - y}{x + y}\right)$$ because multiplying $\theta$ by $(x + y)$ and substituting $\theta$ gives the expression. 10. **Final formula:** Thus, $$P = 4\sqrt{xy} + \pi(x + y) + 2(x - y)\sin^{-1}\left(\frac{x - y}{x + y}\right).$$ This completes the proof.