Let $$a, b$$ be positive integers satisfying $$a^{3}-b^{3}-ab=25$$. Find the largest possible value of $$a^{2}+b^{3}$$.

We start from the given condition

$$a^{3}-b^{3}-ab = 25,\qquad a,b\in\mathbb{Z}_{+}.$$

Because $$a^{3}-b^{3}\gt 0,$$ we must have $$a\gt b.$$ Put $$d=a-b$$ where $$d\in\mathbb{Z}_{+}.$$ Then $$a=b+d$$ and

$$a^{3}-b^{3}=(b+d)^{3}-b^{3}=3b^{2}d+3bd^{2}+d^{3}.$$

Substituting this and $$ab=b(b+d)=b^{2}+bd$$ into the original equation:

$$3b^{2}d+3bd^{2}+d^{3}-(b^{2}+bd)=25.$$

Simplify:

$$(3d-1)(b^{2}+bd)+d^{3}=25.$$

Observe that $$d^{3}\le 25\;\Longrightarrow\;d\le 2,$$ so only $$d=1$$ or $$d=2$$ are possible.

Then $$a=b+1.$$ Compute:

$$a^{3}-b^{3}-(ab)=25\; \Longrightarrow\;(b+1)^{3}-b^{3}-b(b+1)=25.$$

That is

$$\left(3b^{2}+3b+1\right)-(b^{2}+b)=25\;\Longrightarrow\;2b^{2}+2b+1=25.$$

$$2b^{2}+2b-24=0\;\Longrightarrow\;b^{2}+b-12=0.$$

Solving the quadratic, $$b=\dfrac{-1\pm\sqrt{1+48}}{2}=\dfrac{-1\pm7}{2}.$$

Positive integer root: $$b=3.$$ Hence $$a=b+1=4.$$

Then $$a=b+2.$$ The equation becomes

$$(b+2)^{3}-b^{3}-b(b+2)=25.$$

$$6b^{2}+12b+8-(b^{2}+2b)=25\;\Longrightarrow\;5b^{2}+10b+8=25.$$

$$5b^{2}+10b-17=0.$$

The discriminant is $$\Delta=10^{2}-4\cdot5\cdot(-17)=440,$$ whose square root is not an integer, so no positive integer $$b$$ satisfies this equation. Hence case 2 yields no solution.

Therefore the only integer solution is $$a=4,\;b=3.$$

Finally,

$$a^{2}+b^{3}=4^{2}+3^{3}=16+27=43.$$

So the largest possible value of $$a^{2}+b^{3}$$ is 43.