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.