WLOG $a\le b\le c$. Since $c^2$ divides the RHS, it also divides $a^3+b^3+c^3$, hence $c^2\mid a^3+b^3$; in particular, $c^2\le a^3+b^3$. Also, $3c^3 \ge (abc)^2$, so $3c\ge a^2b^2$, which means $c^2 \ge \tfrac{a^4b^4}{9}$. Now, \[ \frac{a^4b^4}{9} \le a^3+b^3 \implies a^4b \le 9\left(\frac{a^3}{b^3}+1\right)\le 9\cdot 2= 18. \]This clearly implies $a=1$. The equation is now $1+b^3+c^3=b^2c^2$. So $2c^3+1 \ge b^2c^2$. If $2c^3+1=b^2c^2$, then $c^2(b^2-2c)=1$, so $c=1$, contradiction. Therefore, $2c^3 \ge b^2c^2$, so $c\ge \tfrac12b^2$. Also, $c^3 \le b^2c^2$, so $c\le b^2$. In summary, $c\in \left[ \tfrac12 b^2, b^2\right]$. Suppose $b\ge 5$. Then $c^2\ge \tfrac14 b^4 \ge b^3+31$. This means $(b^2-c)c^2 \ge b^3+31$ (assuming $c\not = b^2$, but this case is easy to deal with). Hence $b^2c^2 \ge b^3+c^3+31$, contradiction. Therefore, $b\le 4$, and bashing out these cases gives the only solution as $(b,c)=(2,3)$. In conclusion, the answers are $(a,b,c)=(1,2,3)$ and permutations.

The answer is $(1,2,3)$ and permutations which works. Assuming $a \le b \le c$, write the given equation as \[ a^3 + b^3 = c^2(a^2b^2-c). \]Thus actually assume $a \le b \le c < a^2b^2$. Write \begin{align*} 2b^3 &\ge a^3+b^3 = c^2(a^2b^2-c) \\ &\ge \min \left\{ b^2(a^2b^2-b), (a^2b^2-1)^2 \right\} \\ &\ge \min \left\{ b^2(b^2-b), (b^2-1)^2 \right\}. \end{align*}where the second equality follows from the fact $x \mapsto x^2(a^2b^2-x)$ has no local minima for $0 < x < a^2b^2$, hence is minimized at endpoints. This implies $b \le 3$, so we just need to manually examine $1 \le a \le b \le 3$ for solutions in $c$. Doing so gives the solution.

https://artofproblemsolving.com/community/c6h1749156p11418402

https://artofproblemsolving.com/community/c6h359192p1962933

The answers are $(1,2,3)$ and permutations. WLOG let $a\geq b\geq c$. Suppose that $b\geq3$. Define the function $f(t)=t^3-b^2c^2t+b^3+c^3$ so that $a$ is a root of $f$. Observe that: \begin{align*} f(-\infty) &= -\infty < 0 \\ f(0) &= b^3+c^3 > 0 \\ f(b) &= -b^3(bc^2-3)-(b^3-c^3) < 0 \\ f(b^2c^2-1) &= -2b^3(\frac{32}{81}b^2c^4-1)-(\frac{1}{9}b^2c^2-1)(\frac{17}{9}b^2c^2-1)-(b^3-c^3) < 0 \\ f(b^2c^2) &= b^3+c^3 > 0 \end{align*}By the intermediate value theorem, $f$ has a root in each of the intervals $(-\infty,0),(0,b),(b^2c^2-1,b^2c^2)$. Since $f$ is a cubic polynomial, it no other roots. But then $f$ has no integer root at least $b$, which is a contradiction. Thus $b\leq2$. Check that the only solution is $(a,b,c)=(3,2,1)$ as desired.

Without loss of generality say $a\leq b\leq c.$ This is an inequality. Check that $(1,1,c)$ does not work in general to avoid sadness. Note $c^2\mid a^3+b^3,$ so $c^2\leq a^3+b^3.$ Now note that we want \[(abc)^2\leq a^3+b^3+c^3\leq (a^3+b^3)(1+c)\]\[(ab)^2c< 2a^3+2b^3, \text{ since } 1+c>2\]\[a^2b^3<4b^3\]\[a^2<4,\]implying that we want $a<2,$ or in particular, $a=1.$ Now we rinse and repeat. We want \[(bc)^2\leq 1+b^3+c^3\leq (1+b^3)(1+c).\]We claim that equality cannot hold unless $b=2$; note that otherwise $b\mid 1+c$ or $c\geq 2b-1,$ epicly failing for size reasons since this implies $c\geq \frac{3}{2}b,$ and if $b> 2$ and $k=\frac{c}{b}\geq \frac{3}{2},$ \[b^4k^2\geq 1+b^3+bk+b^4k,\text{ since}\]\[bk(b^3k-b^3-1)\leq b^3(k-1)\geq bk\left(\frac{b^3}{2}-1\right)\geq \frac{25}{54}b^4k \text{ and }\]\[b^3+1\leq \frac{28}{27}b^3, \text{ and}\]\[25bk\geq 56,\]contradiction. Thus we want \[(bc)^2\leq b^3c+b^3+c\]\[c(b^2c-1)\leq b^3(c+1)\]and note that $b^2c-1\geq \frac{35}{27}b^3$ as we assume $b\geq 3$ and obviously $b\neq c,$ and that $c+1\leq \frac{5}{4}{c}.$ So we have \[\frac{35}{27}b^3c\leq \frac{5}{4}b^3c,\]which is ridiculous, so we reach a contradiction.

Here's a cool solution I found. Without loss of generality say $a \geqslant b \geqslant c$. We know that $a$ is a root of the polynomial $x^3-(bc)^2x^2+b^3+c^3$. Let $u$ and $v$ be the other two roots. By Vieta's formulas, we have $$a+(u+v)=b^2c^2,$$$$a(u+v)=-uv,$$$$(-uv)a=b^3+c^3.$$From the first equation, we see that $u+v$ is an integer, and from the second and third equation we see that $u+v>0$. Let $w=u+v$. Then, eliminating $uv$, we obtain the following equations: $$a+w=b^2c^2,$$$$aw=\frac{b^3+c^3}{a}.$$Using the inequality $aw+1\geqslant a+w$, we have $$b^2c^2\leqslant \frac{b^3+c^3}{a}+1 \leqslant 2b^2+1 \implies c=1,$$where we used the ordering $a\geqslant b \geqslant c$ for the last inequality. The equations are now $$a+w=b^2,$$$$aw=\frac{b^3+1}{a}\leqslant b^2+1.$$If $w\geqslant 2$ and $a>2$, we have $(a-1)(w-1)\geqslant 2$, which implies $aw>a+w+2 \implies b^2+1>b^2+2$, a contradiction. If $w=1$, then $a=b^2-1$ and $a^2=b^3+1$, which implies $b^2(b^2-2)=b^3 \implies b=2, a=3$, which is a solution. If $a\leqslant 2$, then the only triples we need to check are $(1,1,1)$, $(2,1,1)$ and $(2,2,1)$. Since none of them works, the only solution is $(3,2,1)$.

All answers are the six permutations of $(3,2,1)$. WLOG, let $a\geq b\geq c$, and note that $$(abc)^2\leq 3a^3 \implies a\geq\frac{(bc)^2}{3}.$$On the other hand, $$a^2\mid b^3+c^3\implies a^2\leq 2b^3.$$Combining both bounds gives $\tfrac{(bc)^4}{9}\leq 2b^3\implies bc^4\leq 18$. Thus $c=1$ and $b\leq 18$. Check the remaining cases manually.

Uhh too ez . Similiar to the others but ill post anyways.

Using divisibility twice seems to accelerate the solution, as follows. Let $a\ge b\ge c$, without loss. Then $a^2b^2c^2\le 3a^2$ implying $b^2c^2\le 3a$. Furthermore, $a^2\mid b^3+c^3$then gives $b^3+c^3\ge a^2$. Combining these, $b^3+c^3\ge a^2 \ge \frac{b^4c^4}{9}$. If $c\ge 2$, then $2b^3\ge b^3+c^3 \ge 16b^4/9$, giving $b\le 9/8$, not possible. Thus $c=1$. Now we arrive at $a^3+b^3+1=a^2 b^2$. Next, it is easily seen that $a\ne b$, thus $a^3+b^3+1< 2a^3$. Consequently, $b^2<2a$. Furthermore, $a^2 \mid b^3+1$. Combining these, we have $b^3+1 \ge a^2 >b^4/4$. This inequality is possible only when $2\le b\le 4$. Now, we also have the condition $a^2\mid b^3+1$, and $a>b$. Under this, the cases $b=3,4$ turn out to be impossible. Thus $b=2$. This gives $a=3$. From symmetry, any permutation of $(1,2,3)$ works.

It's really similar to Iran Second Round problem. It was used on a TST in my country and I knew the main solution idea beforehand. https://artofproblemsolving.com/community/c6h1832202p12271047

WLOG $a \ge b \ge c.$ We have $(abc)^2=a^3+b^3+c^3 \le 3a^3 \Longrightarrow (bc)^2 \le 3a.$ But since $a^2 \le b^3+c^3 \Longrightarrow (bc)^2 \le 3\sqrt{b^3+c^3}$ $\Longrightarrow (bc)^4 \le 9(b^3+c^3),$ which fails if both $b, c \ge 3.$ Bash out all possibilities to get $(b,c) = (2,3) \Longrightarrow (a,b,c) = (1,2,3)$

anyone else notice this? shortlist didn't come out in 2018

OlympusHero wrote: anyone else notice this? shortlist didn't come out in 2018 See post #4

The answer is $\boxed{\{a,b,c\}=\{1,2,3\}}$. WLOG $a \leq b \leq c$. Claim: $a=1$ Proof: From the given equation we have $c^2 \vert a^3+b^3$ so: $$c^2 \leq a^3+b^3 \leq 2b^3$$Also we have: $$(abc)^2=a^3+b^3+c^3 \leq 3 c^3 \Rightarrow a^2b^2 \leq 3c \Rightarrow a^6 b^6 \leq 27 c^3$$Combining these two inequalities gives: $$27c^3 \geq a^6 \left(\frac{c^2}{2}\right)^2 \Rightarrow 108=4 \cdot 27 \geq a^6 c \geq a^7$$and as $2^7=128>108$ the claim follows. The equation now reduces to: $$1+b^3+c^3=b^2 c^2 \Leftrightarrow b^2\left(c^2-b\right)=1+c^3 \quad \quad(\blacktriangle)$$This is not solved when $b=c=1$. As $c \geq b$, after excluding this case we have $c^2-b>0$ and: $$c^2-b \vert 1+c^3 \Rightarrow c^2-b \vert 1+c^3-c(c^2-b)=1+bc \quad \quad(\blacksquare)$$\underline{Case 1: $c=b$} In this case the original equation reduces to: $$1=c^4-2c^3=c^2(c-2)$$which forces $c=1$ but this is not a solution. \underline{Case 2: $c \geq b+1$} $(\blacksquare)$ gives: $$1+bc \geq c^2-b \geq c(b+1)-b=bc+(c-b) \geq bc+1$$Hence we must have equality throughout which forces $c=b+1$. Plugging this in to $(\blacktriangle)$ gives: $$(c-1)^2\left(c^2-c+1\right)=1+c^3 \Leftrightarrow (c-1)^2=c+1 \Leftrightarrow c(c-3)=0$$so $c=3$, $b=2$, $a=1$ which is indeed a solution.

Storage y-is-the-best-_ wrote: Find all triples $(a, b, c)$ of positive integers such that $a^3 + b^3 + c^3 = (abc)^2$. WLOG Let $a\geq b\geq c$. Hence, $a^3+b^3+c^3=(abc)^2\le 3a^3\implies (bc)^2\le 3a\implies a\ge \frac{(bc)^2}{3}$. But $a^2\mid a^3+b^3+c^3\implies a^2\mid b^3+c^3\implies a^2\le b^3+c^3\le 2b^3$. Hence, $2b^3\ge \frac{b^4c^4}{9}\implies 18\ge bc^4\implies c=1$ as $c>1$ would lead to $18\le bc^4$. So, $18\ge b$. Now checking the cases when $b=\{1,2,3,\cdots 18\}$. We get that there exists only one solution of $(a,b,c)$ which $(3,2,1)$. Hence, $(a,b,c)=(1,2,3),(1,3,2),(2,3,1),(2,1,3),(3,2,1),(3,1,2)$. $\blacksquare$

amar_04 wrote: N1 was tougher than N2? Storage y-is-the-best-_ wrote: Find all triples $(a, b, c)$ of positive integers such that $a^3 + b^3 + c^3 = (abc)^2$. WLOG Let $a\geq b\geq c$. Hence, $a^3+b^3+c^3=(abc)^2\le 3a^2\implies (bc)^2\le 3a\implies a\ge \frac{(bc)^2}{3}$. But $a^2\mid a^3+b^3+c^3\implies a^2\mid b^3+c^3\implies a^2\le b^3+c^3\le 2b^3$. Hence, $2b^3\ge \frac{b^4c^4}{9}\implies 18\ge bc^4\implies c=1$ as $c>1$ would lead to $18\le bc^4$. So, $18\le b$. Now checking the cases when $b=\{1,2,3,\cdots 18\}$. We get that there exists only one solution of $(a,b,c)$ which $(3,2,1)$. Hence, $(a,b,c)=(1,2,3),(1,3,2),(2,3,1),(2,1,3),(3,2,1),(3,1,2)$. $\blacksquare$ Do you mean $a^3+b^3+c^3=(abc)^2\le 3a^3$? and $18\geq b$

pad wrote: WLOG $a\le b\le c$. Since $c^2$ divides the RHS, it also divides $a^3+b^3+c^3$, hence $c^2\mid a^3+b^3$; in particular, $c^2\le a^3+b^3$. Also, $3c^3 \ge (abc)^2$, so $3c\ge a^2b^2$, which means $c^2 \ge \tfrac{a^4b^4}{9}$. Now, \[ \frac{a^4b^4}{9} \le a^3+b^3 \implies a^4b \le 9\left(\frac{a^3}{b^3}+1\right)\le 9\cdot 2= 18. \]This clearly implies $a=1$. The equation is now $1+b^3+c^3=b^2c^2$. So $2c^3+1 \ge b^2c^2$. If $2c^3+1=b^2c^2$, then $c^2(b^2-2c)=1$, so $c=1$, contradiction. Therefore, $2c^3 \ge b^2c^2$, so $c\ge \tfrac12b^2$. Also, $c^3 \le b^2c^2$, so $c\le b^2$. In summary, $c\in \left[ \tfrac12 b^2, b^2\right]$. Suppose $b\ge 5$. Then $c^2\ge \tfrac14 b^4 \ge b^3+31$. This means $(b^2-c)c^2 \ge b^3+31$ (assuming $c\not = b^2$, but this case is easy to deal with). Hence $b^2c^2 \ge b^3+c^3+31$, contradiction. Therefore, $b\le 4$, and bashing out these cases gives the only solution as $(b,c)=(2,3)$. In conclusion, the answers are $(a,b,c)=(1,2,3)$ and permutations. After demonstrate $c = 1$ we can write. modulo $a$: $a^3+b^3+c^3 = (abc)^2 \implies b\equiv -1 ( \mod a)$ because $a\geq b$ And we get: $a^3+(a-1)^3+ 1 = (a^2 - a)^2 \implies (2a -1)(a^2 - a+1) = (a^2 - a+1)(a^2 - a - 1)$ $\implies a^2 - 3a = 0$ $\implies a = 3 $. And conclusion follow.

This problem sucks. WLOG assume $a \geq b \geq c$. Note that $a^2 \mid a^3 + b^3 + c^3 \implies a^2 \mid b^3 + c^3 \implies a^2 \leq b^3 + c^3$. Furthermore,\[3a^3 \geq a^3 + b^3 + c^3 = a^2b^2c^2 \implies a \geq \tfrac{b^2c^2}{3}.\]Thus, we get ${b^4c^4} \leq 9(b^3 + c^3) \leq 18b^3 \implies bc^4 \leq 18$. Since $a, b, c \in \mathbb{Z}_+$ it follows that $c = 1$ and $b \leq 18$. Doing a manual case check from here on out is not terribly hard; we arrive at solutions $(a, b, c) = (3, 2, 1)$ and its $6$ permutations.

Let $a\ge b\ge c$ then $a^2\mid b^3+c^3$ which implies that $a^2\le b^3+c^3$, but on the other hand $3a^3\ge (abc)^2$ which implies $a^2\ge\frac{b^4c^4}{9}.$ Thus, $18b^3\ge 9(b^3+c^3)\ge b^4c^4$ which implies $\frac{18}{b}\ge c^4$ which means $b=1,c=2$ or $c=1.$ Only the latter applies. $~$ Now, we also see that $b\le 18.$ If $b=2$ then $a=3.$ None of the other cases have any solutions so the answer is $(1,2,3)$ are any permutations.

We claim the answer is permutations of $(1,2,3)$ which obviously work. WLOG, let $a \le b \le c$. We have $a^3+b^3+c^3=(abc)^2 \le 3c^3 \implies a^2b^2 \le 3c$. We also have $c^2|a^3+b^3 \implies c^2 \le a^3+b^3 \le 2b^3$. This means $$a^4b^4 \le 9c^2 \le 18b^3 \implies a^4b \le 18 \implies a^5 \le 18 \implies a=1.$$We now repeat this idea. We have that $1+b^3+c^3=(bc)^2 \le 2c^3 + 1 \implies b^2 \le 2c$. We also have $c^2 |b^3+1 \implies c^2 \le b^3+1$. This means $b^4 \le 4c le 4(b^3+1) \implies b \le 4$. If $b=1$, then $c^2 | 2$, so $c=1$ contradiction. If $b=2$, we have $c^2|9$, so $c=3$ and we get the solution $(1,2,3)$. If $b=3$, then $c^2 | 28 \implies c \le 2$, contradiction. Finally if $b=4$, then $c^2|65$, so $c=1$, contradiction. We have thus checked all cases and so we are done.

WLOG let $a \ge b \ge c$. We have $a^2 \hspace{0.3em} | \hspace{0.3em} a^3+b^3+c^3 \implies a^2 \hspace{0.3em} | \hspace{0.3em} b^3+c^3 \implies a^2 \le b^3+c^3$ On the other hand $(abc)^2 = a^3+b^3+c^3 \le 3a^3 \implies 3a \ge b^2c^2 \implies 9a^2 \ge b^4c^4 \implies 9(b^3+c^3) \ge b^4c^4$. Hence $18b^3 \ge b^4c^4 \implies bc^4 \le 18 \implies c=1 \text{ or } 2$. If $c=2$ then $b=1$, contradicting $b \ge c$. Hence $c=1$ and $9b^3+9 \ge b^4 \implies b \le 10$. Checking each $1 \le b \le 10$ we see that the only solution is $a=1,b=2,c=3$. Hence, the only triples are $(1,2,3)$ and permutations, which clearly work.

WLOG let $a \geq b \geq c$. We have $$3a^3 \geq a^3+b^3+c^3 = a^2b^2c^2 \rightarrow a \geq\frac{b^2c^2}{3}.$$We also have $a^2 \leq 2b^3$ by using $\pmod {a^2}$, and combining yields $bc^4 \leq 18$, which means $c=1$ and $b \leq 18$. Now, manually checking cases will give the only working triple when $b=2$, which is $(1,2,3)$. Then the answer is $(1,2,3)$ and its permutations (since we assumed $a\geq b\geq c$).

WLOG $a \ge b \ge c$. Note that $a^2 | (b^3 + c^3)$ because we can divide by $a^2$ in the original equation and since the RHS is an integer the LHS must be an integer too. So we know that $a^2 \le b^3 + c^3$. By our assumption, $3a^3 \ge a^3 + b^3 + c^3 = (abc)^2$. Hence we know that $a \ge \frac{(bc)^2}{3}$ or $a^2 \ge \frac{b^4c^4}{9}$. This means that \[\frac{b^4c^4}{9} \le b^3 + c^3 \Rightarrow bc^4 \le 18.\]Hence we know that $c= 1$. This now means that $b \le 18$. We can test every value and we quickly see that any permutation of $(1,2,3)$ work.

By the AM-GM inequality, $abc$ is at most $\left(\frac{a+b+c}{3}\right)^3$. So rearranging terms in the inequality, $3^6 \geq \frac{\left(a+b+c\right)^3}{a^3+b^3+c^3}\left(a+b+c\right)^3$. This implies $a+b+c \leq 8$ Now, we can manually check that we cannot have a solution where $2$ of the $3$ numbers are equal, so we must have $(1, 2, 3)$ and $(1, 2, 4)$ and $(1, 3, 4)$ as the only possible solutions out of which only $(1, 2, 3)$ works

Funky solution. Without loss of generality, let $a\geq b \geq c$. It follows that $a^2b^2c^2 \leq 3a^3$, or $b^2c^2 \leq 3a$. Since $a^2 \mid b^3+c^3$, we also have $a^2 \leq b^3+c^3$. From the first inequality, $\frac 19 b^4c^4 \leq a^2$, so \[ \frac 19 b^4c^4 \leq b^3+c^3 \implies 1\leq \frac 9{bc^4} + \frac 9{b^4c} \]If $c\geq 2$, it is easy to check there are no solutions to this inequality. Then $c=1$, and we can have any value of $b$ between $1$ and $9$ inclusive. Checking all of them, we find only $(3,2,1)$ works up to permutation.

The answer is $(a,b,c)=(1,2,3)$ and permutations. As the equation is symmetric, we may assume that $a\ge b\ge c$. First note that $a^2\mid b^3+c^3$, so $a^2\le b^3+c^3.$ Moreover, \[3a^3\ge a^2+b^2+c^2=a^2b^2c^3,\]so $3a\ge b^2c^2$ or $a^2\ge\frac{b^4c^4}{9}$. Combining these two inequalities, \[\frac{b^4c^4}{9}\le b^3+c^3\implies c^4b\le 9\left(1+\frac{b^3}{c^3}\right)\le 18,\]giving $c=1$. Therefore, the equation reduces to \[a^3+b^3+1=a^2b^2.\]Then, as $a^2\le b^3+1$, we have \[a^3+a^2\ge a^3+b^3+1=a^2b^2\implies a+1\ge b^2\implies a^2\ge (b^2-1)^2.\]Putting this together, we have $b^3+1\ge(b^2-1)^2$, which is inductively false for $b>2$. So $b=1$ or $b=2$. Finishing, we see that $b=1$ yields no solution while $b=2$ gives $a=3$.

It's like they made ``annoying size details" into a problem :/ Let $a \leq b \leq c$. Notice that $c^2 \mid a^3+b^3$ and cyclic permutations, which implies that $c \leq \sqrt{a^3+b^3}$. On the other hand, $$3c^3 \geq a^3+b^3+c^3 = a^2b^2c^2 \implies c \geq \frac{a^2b^2}3.$$Now assume $a, b \geq 2$, such that $ab \geq a+b$. Then $$\frac{a^4b^4}9 \geq \frac{(a+b)^4}9 > \frac{a^4+4a^3b+4ab^3+b^4}9 > a^3+b^3$$as $a+4b \geq 10 > 9$ and $4a+ b>9$. So we may assume $a = 1$, from where the given equation becomes $\frac{b^4}9 < b^3+1$. We need to check $b \in \{1, 2, \dots, 9\}$, which yields the only solution triple $(1, 2, 3)$ and cyclic permutations.

WLOG let, $a \geq b \geq c$. We have $a^2 \mid b^3 + c^3 \implies a^2 \leq b^3 + c^3.$ On the other hand we have $3a^3 \geq (abc)^2 \implies a \geq \frac{(bc)^2}{3}.$ Putting all of them together, we see \[\frac{(bc)^4}{9} \leq b^3 + c^3 \implies \frac{bc^4}{9} \leq 1 + \left(\frac{c}{b}\right)^3 \leq 2.\] Hence, $bc^4 \leq 18$ which gives $c = 1$. Thus our equation becomes $a^2 + b^2 + 1 = (ab)^3.$ Note that $a \neq 1$ because $(1,1,1)$ is not a solution. Hence $a > 1.$ Proceeding similarly, we have $ 2a^3 \geq (ab)^2 - 1 \implies 2a \geq {b^2} - \frac{1}{a^2} > b^2 - 1.$ This implies $a \geq \frac{b^2}{2}.$ But as, $a^2 \leq b^3 + 1 \implies \frac{b^4}{4} \leq b^3 + 1.$ This gives us that $b \leq 4.$ Checking each cases manually we find that only $b = 2$ works. This gives us $a = 3.$ Hence the solutions are $(3,2,1)$ and permutations.

Solved with ihategeo_1969. We found it funny that this is actually the base case of 2019 Iran MO Round 2 Problem 3 but I guess there's probably a perfectly valid explanation for this. We claim that the only triple of solutions (upto permutations) is $(1,2,3)$. The equations is entirely symmetric so we assume $a\ge b\ge c$. Now, since $a^2$ is a factor of the left hand side it must also be a factor of the right hand side, so $a^2\mid b^3+c^3$. Then, $a^2\le b^3+c^3$. But, note that, \[3a^3 \ge a^3+b^3+c^3 = (abc)^2\]So, $3a \ge b^2c^2$. Combining this with our previous inequality we have that if $c>1$, \[18c^3 \ge 9(b^3+c^3)\ge 9a^2 \ge (b^2c^2)^2 = b^4c^4 \ge 16c^4 \ge 32c^3\]which is a very clear contradiction since $c$ is a positive integer. Thus, $c=1$. Repeating the bounding with this additional piece of information, \[18=18c^3 \ge 9(b^3+c^3)\ge 9a^2 \ge (b^2c^2)^2 = b^4c^4 = b^4\]so $b=1$ or $b=2$. Of these, when $b=1$ we have $a^2=a^3+2$ which quite clearly has no solutions, and when $b=2$ we have $4a^2 = a^3+9$ which implies $a\le 4$ of which only $a=3$ work giving us our desired solution set.

2019 N2 We claim that the answer is $(3,2,1)$ upto permutations. Assume WLOG $a \geq b \geq c$, then we have that $3a \ge b^2c^2$, and $ a^2 | b^3+c^3$. Combining these we get that $c=1$. Now our equation is. \[a^3+b^3+1=(ab)^2\]This implies that $a^2 | b^3+1 = (b+1)(b^2+1-b)$. But we have that $a^2 > b+1$ and $a^2 \geq b^2+1-b$, where equality is when $b=a=1$, hence $a^2 > b^2+1-b$, implying that $a^2=b^3+1$, subtracting one we get that. \[(a-1)(a+1)= b^3\] For the other case, if $a | b+1$ and $a| b^2+1-b$ this implies that $a=b+1$ and that $b+1 | b^2+1-b.$ Which holds when $a=3$ and $b=2$. Hence we are done

WLOG $a \le b \le c$. Note that $(abc)^2 \le 3c^3 \implies a^2b^2 \le 3c$, and $c^2 \mid a^3+b^3 \implies 9(a^3+b^3) \ge a^4+b^4$. Analysing $f(a, b) = a^4+b^4 - 9 (a^3+b^3)$, we see it is increasing in both $a, b \in \mathbb{N}$ and positive for $(2, 2)$, so in fact $a = 1$. From here similar arguments yield $b^4 \le 4b^3+4 \implies b \le 4$, and case bashing gives $(a, b, c) = (1, 2, 3)$ and permutations.

nice Only two observations are needed. Assume $a\le b\le c$. Then we get: First, notice $(abc)^2=a^3+b^3+c^3\le 3c^3\implies (ab)^2\le 3c$. Second, notice $a^3+b^3\equiv a^3+b^3+c^3=(abc)^2\equiv 0\pmod{c^2}$. Hence write $k:=\frac{a^3+b^3}{c^2}$. Notice that \[a^3+b^3\le 1^3+(ab)^3\le 1+(3c)^{3/2}\implies kc^2\le 1+(3c)^{3/2}\implies k\le \frac{1}{c^2}+\sqrt{\frac{27}{c}}.\]From here we simply bound: notice that $c\ge 28$ fails, as we obtain \[1\le k\le \frac{1}{28^2}+\sqrt{\frac{27}{28}}\implies \frac{28^2-1}{28^2}\le \sqrt{\frac{27}{28}}\implies \left(\frac{28^2-1}{28^2}\right)^2\le \frac{27}{28}\implies 29\cdot 27\cdot 29\le 28\cdot 28\cdot 28\]or if $x:=28$ then $x^3+x^2-x-1=(x-1)(x+1)^2\le x^3$ which is false. Thus $1\le c\le 27$. Notice now that $a^3+b^3\le 1+(3c)^{3/2}\le 730\implies 1\le b\le 9$. We split into cases: If $b=1$, then $a=1$---and thus $c^2\mid 2$, so that $a=b=c=1$, which fails. If $b=2$, then $a=1$ or $a=2$. In the first case, we find $c^2\mid 9$, so that $c=3$. The solution $(1,2,3)$ works. In the second case, we find $c^2\mid 16$, so that $c=2$ or $c=4$, neither of which work. If $b=3$, then $a=1$, $a=2$, or $a=3$, so that $a^3+b^3\in \{28,35,54\}$. In this case, only $c=3$ is valid, from $c^2\mid 54$. This fails the original equation. If $b=4$, then $a\in \{1,2,3,4\}$, so that $a^3+b^3\in \{65,72,91,128\}$. These numbers yield $c=6$ from $c^2\mid 72$ and $c\in {4,8}$ from $c^2\mid 128$. None of these work---the check, by the way, is being done by checking if $a^3+b^3+c^3$ is a square. If $b=5$, then $a\in \{1,2,3,4,5\}$, so that $a^3+b^3\in \{126,133,152,189,250\}$. Only $c=5$ works from $c^2\mid 250$, and this fails. If $b=6$, then $a\in \{1,2,3,4,5,6\}$, so that $a^3+b^3\in \{217,224,243,280,341,432\}$. Only $c=9$ works from $c^2\mid 243$, as well as $c\in \{6,12\}$ from $c^2\mid 432$. None of these work. If $b=7$, then $a\in \{1,2,3,4,5,6,7\}$, so that $a^3+b^3\in \{344,351,370,407,468,559,686\}$. Only $c=7$ works from $c^2\mid 686$, and this fails. If $b=8$, then $a\in \{1,2,3,4,5,6\}$ from $a^3+b^3\le 730$, so that $a^3+b^3\in \{513,520,539,576,637,728\}$. From here, only $c\in\{8,12,24\}$ work from $c^2\mid 576$. These each fail the original equation. If $b=9$, then $a=1$ from $a^3+b^3\le 730$, and no value works for $c^2\mid 730$. We have exhausted all cases. The only working triplet is $(1,2,3)$ and its permutations.