<s>[INST] How find this diophantine equation integer solution $a^3+b^3=(2ab+1)^2$ 
Find this following diophantine equation  integer solution 
  $$a^3+b^3=(2ab+1)^2$$

I think this equation only have two following solution
$$(a,b)=(1,0),(0,1)$$ 
maybe this equation have no other solution? because can see wolframalpha.com
I think following idea is usefull
$$a^3+b^3=(a+b)(a^2-ab+b^2)=(2ab+1)^2$$
since
$(a+b,a^2-ab+b^2)=1$ so
$a+b,a^2-ab+b^2$ are square numbers?
 [/INST] Let's rewrite equation in $s,t$ doetoe (edit: he deleted his answer, look at the end of mine) got as follows: $4t^2+(3s+4)t+(1-s^3)=0$, and try to solve it for $t$. Because $t$ is integer, we must get that discriminant is a perfect square: $\Delta=(3s+4)^2-4\cdot 4(1-s^3)=9s^2+24s+16-16+16s^3=s(16s^2+9s+24)$. It's easy to see that, assuming $a,b$ satisfy OP's equation, that $2\nmid s$, and we can check that $3\nmid s$: if $3\mid a,b$ then $3\mid a^3+b^3$ but $3\nmid (2ab+1)$, and if $a\equiv 1,b\equiv -1\mod 3$ then again $3\mid a^3+b^3$ but $3\nmid (2ab+1)$. So $s$ and $24$ are relatively prime, and thus $s$ and $16s^2+9s+24$ are, so, because these two multiply to a perfect square, both must be perfect squares. But $16s^2+9s+24=(4s+1)^2+s+23$. Now, for $s>\frac{20}{7}: (4s+1)^2<(4s+1)^2+s+23<(4s+2)^2$ and for $s<-23: (4s+1)^2>(4s+1)^2+s+23>(4s+2)^2$ (simple quadrtatic inequalities, can be checked with Wolfram Alpha). So, unless $-23\leq s\leq \frac{20}{7}$ $\Delta$ cannot be perfect square. One can check that in this interval $\Delta$ is square for $s=1$ or $s=-23$. For $s=1$ we get $t=0$ or $-\frac{7}{4}$. First leads to solutions you have found, second doesn't give integer solutions. $s=-23$ doesn't result in real solutions.
So only solutions are indeed $(0,1),(1,0)$.
EDIT: doetoe's answer has expressed OP's equality in terms of $s=a+b,t=ab$ and it wad form $4t^2+3ts+4t=s^3-1$, and equivalence can be easily verified by substitution and elementary transformations.
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