Diophantine equations - Perfect square and Perfect cube related Solve following Diophantine equations:
$1) \ a^3-a^2+8=b^2$
2) $a, \ b,\  c \in \mathbb{Z^+}$$$\frac{a^3}{(b+3)(c+3)} + \frac{b^3}{(c+3)(a+3)} + \frac{c^3}{(a+3)(b+3)} = 7$$
3) $a^3-8=b^2$
In Problem 2 I tried to use inequality, then I can 'limit' that: $25 \ge a+b+c$ and $a^3 + b^3 + c^3 \ge 112$
Please use elementary way to solve it, I haven't studied elliptic curve yet, thanks.

(2) Extremely ugly solution.
You have
$$a^3(a+3)+b^3(b+3)+c^3(c+3)=7(a+3)(b+3)(c+3)$$
or
$$(x-3)^3x+(y-3)^3y+(z-3)^3z=7xyz$$
Since it is symmetric, we can look for the solutions where $x \geq y \geq z$.
It is easy to check that for $x>15$ we have $(x-3)^3>7x^2$. This shows that $4 \leq x \leq 14$. For each particular $x$ you get a simpler solution which can be solved the same way.
P.S. It probably also helps observing that modulo 3 you have 
$$(w-3)^3w \equiv 0,1 \pmod 3$$
Then, if none of $x,y,z$ is divisible by $3$ the LHS is 0 $\pmod 3$ which is not possible.
If one of $x,y,z$ is divisible by $3$, then all of them must be divisible by 3, and looking at the equation it follows that one of them is divisible by 9. 
This should solve the equation. 
