Solution: Substitute $ x = 8 $: $ S = \frac{6}{\sqrt{8}} = \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}} $. Rationalize: $ \frac{3\sqrt{2}}{2} $. Final answer: $ \boxed{\dfrac{3\sqrt{2}}{2}} $.

["Substituting $ x = 8 $ and Rationalizing $ \frac{6}{\sqrt{8}} $: A Step-by-Step Solution", "When solving expressions involving square roots, especially in algebraic equations, accurate substitution and rationalization are essential for clarity and simplification. In this guide, we explore a key calculation: substituting $ x = 8 $ into a simplified expression derived from such a context, demonstrating how to rationalize a radical expression to arrive at its simplest form.", "---", "### The Expression: $ \frac{6}{\sqrt{8}} $", "At first glance, $ \frac{6}{\sqrt{8}} $ appears straightforward, but working with radicals in equations—particularly when substituting values—requires rationalizing the denominator. This process eliminates irrational numbers from the denominator, making subsequent calculations cleaner and more precise.", "---", "### Step 1: Simplify the Square Root in the Denominator", "Begin by simplifying $ \sqrt{8} $. Since $ 8 = 4 \ imes 2 $, and $ \sqrt{4} = 2 $, we rewrite:", "$$\n\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}\n$$", "Thus, the expression becomes:", "$$\n\frac{6}{\sqrt{8}} = \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}}\n$$", "---", "### Step 2: Rationalize the Denominator", "To rationalize $ \frac{3}{\sqrt{2}} $, multiply both numerator and denominator by $ \sqrt{2} $—the conjugate denominator:", "$$\n\frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\n$$", "This step ensures the denominator is a rational number, complying with standard algebraic conventions for simplifying radical expressions.", "---", "### Final Result", "After substitution and rationalization, the fully simplified value is:", "$$\n\boxed{\dfrac{3\sqrt{2}}{2}}\n$$", "---", "Why Rationalization Matters\nRationalizing denominators not only simplifies expressions but also supports further algebraic manipulation, comparison, and computation—particularly in advanced math, physics, and engineering applications. It ensures clarity and precision when presenting results.", "---", "Example Summary\nStart with:\n$$\n\frac{6}{\sqrt{8}}\n\Rightarrow\n\frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}}\n\Rightarrow\n\frac{3\sqrt{2}}{2}\n$$\nFinal answer:\n$$\n\boxed{\dfrac{3\sqrt{2}}{2}}\n$$", "This step-by-step method ensures every student and practitioner can confidently handle radicals in equations and real-world problems."]