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The volume formula for a sphere is given by \(V = \frac{4}{3} \pi r^3\). This equation helps us determine the volume \(V\) of a sphere when we know its radius \(r\). Spheres are 3-dimensional shapes where every point on the surface is equidistant from the center. To apply the volume formula, you need to correctly use the sphere's radius.

Thus, \(V\) is directly proportional to the cube of the radius \(r\). Larger radii result in larger volumes. The factor \(\frac{4}{3}\pi\) is a constant that relates these two quantities.

For example, doubling the radius increases the volume by a factor of eight (since \(2^3 = 8\)). Conversely, halving the radius reduces the volume to one-eighth of the original.

Let's look at how you can isolate and find exact values using this equation.

To solve for \(r\) in the equation \(V = \frac{4}{3}\pi r^3\), start by isolating the variable \(r\). This involves rearranging the equation so that \(r^3\) is by itself on one side. This is done in several steps:

• First, multiply both sides of the equation by the reciprocal of the coefficient in front of \(r^3\). Here, multiply by \(\frac{3}{4\pi}\).
• This makes the equation simpler because the constants cancel each other out, leaving only \(r^3\) isolated on one side.

After this step, you get: \[\frac{3V}{4\pi} = r^3\].

These intermediary steps are crucial to understand. They highlight the importance of performing algebraic operations to isolate variables, a fundamental procedure in algebra.

To finally solve for \(r\), you need to take the cube root of both sides of the equation \(\frac{3V}{4\pi} = r^3\).

A cube root is a number that, when multiplied by itself three times, gives the original number. For instance, the cube root of 27 is 3 because \(3 \times 3 \times 3 = 27\).

In mathematical terms, taking the cube root of a number \(x\) is denoted as \(x^{1/3}\) or \sqrt[3]{x}\.

Applying this to our equation \[ r = \sqrt[3]{\frac{3V}{4\pi}} \] will give you the value of \(r\).

Understanding cube roots is essential as it allows you to move from \(r^3\) back to \(r\), making it an invaluable tool in algebra.