7. $a, b, c$ are natural numbers. (1) Find the solutions $(a, b)$ that satisfy $a>b$ and $(a+b)^{2}=a^{3}+b^{3}$, (2) Find the solutions $(a, b, c)$ that satisfy $a>b>c$ and $(a+b+c)^{2}=a^{3}+b^{3}+c^{3}$.
7. (1) Answer: $(a, b)=(2,1)$.
Hint: $a^{3}+b^{3}-(a+b)^{2}=(a+b)$
$[a(a-b-1)+b(b-1)] \geqslant 0$, thus $a^{3}$
$+b^{3} \geqslant(a+b)^{2}$, equality holds if and only if $a-b-1=0$,
$b-1=0$, so $(a, b)=(2,1)$.
(2) Answer: $(a, b, c)=(3,2,1)$.
Hint: $a^{3}+b^{3}+c^{3}-(a+b+c)^{2}$ $=a\left[(a-3)^{2}+5(a-b-1)+3(b-c-\right.$ 1) $+(c-1)]+\left[b^{3}+c^{3}-(b+c)^{2}\right]=0$. From (1), we know $b^{3}+c^{3}-(b+c)^{2}=0$, and since $a, b$, $c$ are natural numbers, and $a>b>c$, the equality holds if and only if $a-3=0, a-b-1=0$, $b-c-1=0, c-1=0$.