To solve the equation:

\[
\frac{1}{6b^2} + \frac{1}{6b} = \frac{1}{b^2}
\]

let's simplify and solve for \( b \):

### Step 1: Find a common denominator

For the fractions on the left side, the common denominator is \( 6b^2 \). Rewrite each term with this common denominator:

\[
\frac{1}{6b^2} + \frac{b}{6b^2} = \frac{1 + b}{6b^2}
\]

So, the equation becomes:

\[
\frac{1 + b}{6b^2} = \frac{1}{b^2}
\]

### Step 2: Eliminate the fractions by multiplying by \( 6b^2 \)

Multiply both sides by \( 6b^2 \) to clear the fractions:

\[
1 + b = 6
\]

### Step 3: Solve for \( b \)

Subtract 1 from both sides:

\[
b = 5
\]

So, the solution is:

\[
b = 5
\]

Would you like further details on the solution or have any questions?

### Related Questions:

1. What if the equation was \(\frac{1}{6b^2} + \frac{1}{6b} = \frac{1}{3b^2}\)?
2. How would you solve \(\frac{1}{ab^2} + \frac{1}{ab} = \frac{1}{b^2}\) where \( a \neq 6 \)?
3. Can you find \( b \) if the equation is modified to \(\frac{1}{6b^3} + \frac{1}{6b} = \frac{1}{b^3}\)?
4. What is the behavior of the function \(\frac{1}{6b^2} + \frac{1}{6b} - \frac{1}{b^2}\) as \( b \) approaches infinity?
5. How would this solution change if the equation was given in terms of decimals instead of fractions?

**Tip:** When equations have fractions, multiplying by the common denominator is a great way to simplify!

Learn how to solve the fractional equation 1/(6b^2) + 1/(6b) = 1/(b^2) step-by-step. This problem involves solving a complex equation with fractions and requires understanding of equations and algebraic manipulation. Suitable for advanced high school students and beyond.

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