Proof verification of a question associated with perfect cubes

Question

(Poland) For positive integers $a \leq b$, do the following items:
(a) Show that $b^3 < b^3 + 6ab + 1 < (b+2)^3$.
(b) Find all such a and b for which both $a^3 + 6ab +1 $ and $b^3 + 6ab + 1$ are perfect cubes.
Proof: (a) We see that $$8+6b(b+2)\geq 8+6b(a+2)>8+6ab>1+6ab>0$$
I just add a $b^3$ term to each of these to get given inequality.
(b) We see from (a) that
$$b^3 < b^3 + 6ab + 1 < (b+2)^3$$
Thus, the only possible value of $b^3+6ab+1$ for it to be a cube is $(b+1)^3$ and thus,$$6ab=3b(b+1)\Rightarrow 2a-1=b$$ $$a^3+6ab+1-(a+1)^3=6ab-3a(a+1)=6a(2a-1)-3a(a+1)=9a(a-1)\geq 0$$and $$(a+2)^3-(a^3+6ab+1)=7+6a(a+2)+6a(2a-1)>0$$So for $a^3+6ab+1$ to be a perfect cube,
the only way left is by letting $ a^3+6ab+1-(a+1)^3=0$,i.e,$9a(a-1)=0\Rightarrow a=1$ and consequently $b=1$ by putting the obtained vale of $a$ in the equation $2a-1=b$.

Answer 1

In (b), I get using $b=2a-1$, $$\begin{align}(a+2)^3-(a^3+6ab+1)&=(a^3+6a^2+12a+8)-(a^3+12a^2-6a+1) \\&=-6a^2+18a-7)\\&=-6a(a-3)-7 \end{align}$$ and this is negative for $a\ge4$.

In fact, $a^3+6ab+1=a^3+12a^2-6a+1$ will rather be between $(a+3)^3$ and $(a+4)^3$ except for small values of $a$.