V_1 = \frac{4}{3}\pi (2x)^3 = \frac{4}{3}\pi (8x^3) = \frac{32}{3}\pi x^3

["Understanding the Volume Formula: V₁ = (4/3)π(2x)³ – A Step-by-Step Breakdown", "When calculating the volume of a sphere, the standard formula is:", "[\nV = \frac{4}{3}\pi r^3\n]", "where ( r ) is the radius of the sphere. In some problems, the radius isn’t directly given in simple terms but is expressed as a multiple of a variable — such as ( r = 2x ). This results in a slightly more complex yet elegant derivation, leading to the volume expression:", "[\nV_1 = \frac{4}{3}\pi (2x)^3 = \frac{4}{3}\pi (8x^3) = \frac{32}{3}\pi x^3\n]", "### What Does This Equation Represent?", "The expression\n[\nV_1 = \frac{32}{3}\pi x^3\n]\nshows the precise volume of a sphere with radius ( 2x ), where ( x ) is a positive scaling variable. Using ( 2x ) inside the radius formula scales the volume accordingly — since volume depends on the cube of the radius, doubling the radius increases the volume by a factor of ( 2^3 = 8 ), multiplied by the original volume factor ( \frac{4}{3}\pi ).", "### Step-by-Step Derivation", "1. Start with the volume formula for a sphere:\n [\n V = \frac{4}{3}\pi r^3\n ]", "2. Substitute radius ( r = 2x ):\n [\n V_1 = \frac{4}{3}\pi (2x)^3\n ]", "3. Evaluate ( (2x)^3 ):\n [\n (2x)^3 = 2^3 \cdot x^3 = 8x^3\n ]", "4. Plug back into the volume formula:\n [\n V_1 = \frac{4}{3}\pi \cdot 8x^3 = \frac{32}{3}\pi x^3\n ]", "### Why This Formula Matters", "Understanding how scaling the radius affects volume is essential in geometry, engineering, physics, and design. It helps visualize how small changes in size yield significant differences in capacity — crucial for tanks, spheres in space, or even biological modeling.", "### Final Summary", "The formula\n[\nV_1 = \frac{32}{3}\pi x^3\n]\nis the expanded form of the volume of a sphere with radius ( 2x ). By cube-expanding ( (2x) ), we reveal how volume scales with dimension, reinforcing fundamental concepts in mathematics and applied sciences.", "---", "Key Takeaways:\n- Volume of sphere: ( \frac{4}{3}\pi r^3 )\n- Scaling radius by 2 increases volume by factor of 8\n- Substituting ( r = 2x ) yields ( V = \frac{32}{3}\pi x^3 )\n- Useful for dimensional analysis and real-world applications", "This formula is a clear example of how algebra simplifies geometric calculations, accelerating problem-solving in STEM fields."]