We are asked to simplify the expression:

\[
\left(\frac{a}{4} - \frac{b}{6}\right)^3 + \left(\frac{a}{4} + \frac{b}{6}\right)^3
\]

This is a sum of cubes of binomials. We can apply the **identity** for the sum of cubes:

\[
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
\]

Let:
\[
x = \frac{a}{4} - \frac{b}{6} \quad \text{and} \quad y = \frac{a}{4} + \frac{b}{6}
\]

We now apply the identity:

\[
x^3 + y^3 = \left[\left(\frac{a}{4} - \frac{b}{6}\right) + \left(\frac{a}{4} + \frac{b}{6}\right)\right] \left[\left(\frac{a}{4} - \frac{b}{6}\right)^2 - \left(\frac{a}{4} - \frac{b}{6}\right)\left(\frac{a}{4} + \frac{b}{6}\right) + \left(\frac{a}{4} + \frac{b}{6}\right)^2\right]
\]

### Step 1: Simplifying the sum \( x + y \)
\[
\left(\frac{a}{4} - \frac{b}{6}\right) + \left(\frac{a}{4} + \frac{b}{6}\right) = \frac{a}{4} + \frac{a}{4} = \frac{2a}{4} = \frac{a}{2}
\]

### Step 2: Simplifying the second factor
We will now simplify the second factor:

1. \( \left(\frac{a}{4} - \frac{b}{6}\right)^2 \)
\[
= \left(\frac{a}{4}\right)^2 - 2 \cdot \frac{a}{4} \cdot \frac{b}{6} + \left(\frac{b}{6}\right)^2 = \frac{a^2}{16} - \frac{ab}{12} + \frac{b^2}{36}
\]

2. \( \left(\frac{a}{4} - \frac{b}{6}\right)\left(\frac{a}{4} + \frac{b}{6}\right) \)
\[
= \left(\frac{a}{4}\right)^2 - \left(\frac{b}{6}\right)^2 = \frac{a^2}{16} - \frac{b^2}{36}
\]

3. \( \left(\frac{a}{4} + \frac{b}{6}\right)^2 \)
\[
= \left(\frac{a}{4}\right)^2 + 2 \cdot \frac{a}{4} \cdot \frac{b}{6} + \left(\frac{b}{6}\right)^2 = \frac{a^2}{16} + \frac{ab}{12} + \frac{b^2}{36}
\]

### Step 3: Putting it all together
Now we combine everything into the full expression:

\[
x^3 + y^3 = \frac{a}{2} \left[ \frac{a^2}{16} - \frac{ab}{12} + \frac{b^2}{36} - \left(\frac{a^2}{16} - \frac{b^2}{36}\right) + \frac{a^2}{16} + \frac{ab}{12} + \frac{b^2}{36} \right]
\]

Simplify the terms inside the bracket:

- The \( ab \) terms cancel out.
- Combine the \( a^2 \) terms: \( \frac{a^2}{16} + \frac{a^2}{16} - \frac{a^2}{16} = \frac{a^2}{16} \)
- Combine the \( b^2 \) terms: \( \frac{b^2}{36} + \frac{b^2}{36} - \frac{b^2}{36} = \frac{b^2}{36} \)

Thus, the expression simplifies to:

\[
x^3 + y^3 = \frac{a}{2} \left( \frac{a^2}{16} + \frac{b^2}{36} \right)
\]

Finally, distribute the \( \frac{a}{2} \):

\[
= \frac{a}{2} \cdot \frac{a^2}{16} + \frac{a}{2} \cdot \frac{b^2}{36} = \frac{a^3}{32} + \frac{ab^2}{72}
\]

### Final simplified expression:
\[
\left(\frac{a}{4} - \frac{b}{6}\right)^3 + \left(\frac{a}{4} + \frac{b}{6}\right)^3 = \frac{a^3}{32} + \frac{ab^2}{72}
\]

Would you like further details or have any questions?

Here are 5 related questions for further practice:
1. Simplify \( (x - y)^3 + (x + y)^3 \).
2. Expand and simplify \( (2a - 3b)^3 \).
3. What is the sum of cubes identity?
4. Simplify \( \left(\frac{a}{2} - \frac{b}{3}\right)^3 + \left(\frac{a}{2} + \frac{b}{3}\right)^3 \).
5. Expand \( (a + b)^3 \).

**Tip**: In algebra, using identities can significantly simplify complex expressions.

Discover how to simplify the cubic binomial expression (a/4 - b/6)^3 + (a/4 + b/6)^3 using the sum of cubes identity. This algebraic problem involves applying key polynomial and cubic identities. Ideal for students in Grades 10-12.

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