The expression in the image is:

$\left( \frac{1}{3}a + \frac{2}{3}b \right)^3 - \left( \frac{1}{3}a - \frac{2}{3}b \right)^3$

We are tasked with simplifying this expression. This is a difference of cubes, and we can apply the identity for the difference of cubes:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

Let:

$x = \frac{1}{3}a + \frac{2}{3}b \quad \text{and} \quad y = \frac{1}{3}a - \frac{2}{3}b$

Thus, we can apply the difference of cubes formula:

1. Subtract the two terms in parentheses:

$x - y = \left( \frac{1}{3}a + \frac{2}{3}b \right) - \left( \frac{1}{3}a - \frac{2}{3}b \right)$

Simplifying:

$x - y = \frac{1}{3}a + \frac{2}{3}b - \frac{1}{3}a + \frac{2}{3}b = \frac{4}{3}b$

2. Compute $x^2 + xy + y^2$. We'll calculate each part separately.

• $x^2 = \left( \frac{1}{3}a + \frac{2}{3}b \right)^2 = \frac{1}{9}a^2 + \frac{4}{9}ab + \frac{4}{9}b^2$
• $y^2 = \left( \frac{1}{3}a - \frac{2}{3}b \right)^2 = \frac{1}{9}a^2 - \frac{4}{9}ab + \frac{4}{9}b^2$
• $xy = \left( \frac{1}{3}a + \frac{2}{3}b \right) \left( \frac{1}{3}a - \frac{2}{3}b \right) = \frac{1}{9}a^2 - \frac{4}{9}b^2$

Adding these together:

$x^2 + xy + y^2 = \left( \frac{1}{9}a^2 + \frac{4}{9}ab + \frac{4}{9}b^2 \right) + \left( \frac{1}{9}a^2 - \frac{4}{9}b^2 \right) + \left( \frac{1}{9}a^2 - \frac{4}{9}ab + \frac{4}{9}b^2 \right)$

Simplifying:

$x^2 + xy + y^2 = \frac{3}{9}a^2 + \frac{8}{9}b^2$

Therefore, the fully simplified expression becomes:

$\frac{4}{3}b \times \left( \frac{3}{9}a^2 + \frac{8}{9}b^2 \right) = \frac{4}{3}b \times \frac{1}{3}(a^2 + 8b^2) = \frac{4}{9}b(a^2 + 8b^2)$

The final simplified form is:

$\frac{4}{9}b(a^2 + 8b^2)$

Would you like further details on any step?

Here are five related questions:

1. How would this formula change if the exponents were different (e.g., squared instead of cubed)?
2. What is the difference between a sum of cubes and a difference of cubes?
3. How can the binomial expansion help in expanding powers of sums or differences?
4. What happens if $a = b$ in this particular expression?
5. How does factoring help simplify cubic expressions?

Tip: When simplifying expressions involving cubes, look for patterns such as the difference or sum of cubes to apply special identities.