1. **State the problem:** Given $a = \frac{16}{3}$ and $b = \frac{7}{3}$, find the value of $\frac{a^3 + b^3}{a + b} \times ab$. 2. **Recall the algebraic identity:** The sum of cubes factorization is: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ 3. **Simplify the expression using the identity:** $$\frac{a^3 + b^3}{a + b} = a^2 - ab + b^2$$ Therefore the original expression becomes $$\left(a^2 - ab + b^2 \right) \times ab$$ 4. **Calculate each element:** - Calculate $a + b$ (not needed explicitly after step 3). - Calculate $ab = \frac{16}{3} \times \frac{7}{3} = \frac{112}{9}$. - Calculate $a^2 = \left(\frac{16}{3}\right)^2 = \frac{256}{9}$. - Calculate $b^2 = \left(\frac{7}{3}\right)^2 = \frac{49}{9}$. 5. **Substitute into $a^2 - ab + b^2$:** $$\frac{256}{9} - \frac{112}{9} + \frac{49}{9} = \frac{256 - 112 + 49}{9} = \frac{193}{9}$$ 6. **Multiply by $ab$:** $$\frac{193}{9} \times \frac{112}{9} = \frac{193 \times 112}{81} = \frac{21616}{81}$$ 7. **Simplify the fraction if possible:** Both numerator and denominator divisible by 1 only, so final simplified value is $$\frac{21616}{81}$$ **Final answer:** $$\frac{21616}{81}$$