Unique Parts

Since $a + b = 15$ we can write the given function as $X(a) = a^{3}(15 - a)^{2}.$ To find any critical points we set $\dfrac{dX}{da} = 0.$ Using the product rule, we find that

$\dfrac{dX}{da} = 3a^{2}(15 - a)^{2} + a^{3}*2(15 - a)*(-1) = a^{2}(15 - a)(3*(15 - a) - 2a) =$

$a^{2}(15 - a)(45 - 5a) = 0$ when $a = 0, 15$ and $9.$

Now since $X(0) = X(15) = 0$ and $X(9) = 9^{3}(15 - 9)^{2} = 26244$ we can conclude that $X$ has a maximum value of $\boxed{26244}$ at $a = 9, b = 6.$

We could also use the AM-GM inequality, (since $a$ and $b$ are both non-negative). Then

$a + b = \dfrac{a}{3} + \dfrac{a}{3} + \dfrac{a}{3} + \dfrac{b}{2} + \dfrac{b}{2} \ge 5\sqrt[5]{\dfrac{a^{3}b^{2}}{3^{3}2^{2}}} \Longrightarrow \left(\dfrac{15}{5}\right)^{5} \ge \dfrac{a^{3}b^{2}}{3^{3}2^{2}} \Longrightarrow a^{3}b^{2} \le 3^{8}2^{2} = 26244,$

with equality holding for $a^{3} = 3^{6}, b^{2} = 3^{2}2^{2} \Longrightarrow a = 9, b = 6.$