Problems & Solutions on Fermat Theorem of Multiple of 3 I am working on an assignment in elementary number theory, in which I have to come up with original problems and then work out their solutions on Fermat theorem of multiple of 3, that is, the equation
$$x^3 + y^3 = z^3$$
does not have integer solution for $xyz \neq 0$.
The first problem that I came up easily is one requiring readers to prove that $x^6 + y^6 = z^6$ does not have integer solution, which is very straightforward. But I need more than just one problem. I have searched around in the internet for ideas but could not find any. Therefore here is my question:
Do you have any ideas, suggestion, links or hints that perhaps can help me with writing couple more question related to the subject? Thank you for your time and help.

Solve in integers $\left(a-b\right)\left(a^2-b^2\right)\left(a^3-b^3\right)=3c^3$.
To solve it, notice $(3c)^3=(a^2-2b^2+ab)^3 + (2a^2-b^2-ab)^3$, 
so either $c=0$ or $a^2-2b^2+ab=0$ or $2a^2-b^2-ab=0$.
Here's where I found the problem.

$$(a-b)\left(a^2-b^2\right)\left(a^3-b^3\right)=(a-b)^3(a+b)\left(a^2+ab+b^2\right)$$
$a^2+ab+b^2=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge 0$ with equality if and only if $a=b=0$.
If $c=0$, then either $a=b$ or $a=-b$, so $(a,b,c)=(k,k,0),(k,-k,0)$, $k\in\mathbb Z$.
If $a^2-2b^2+ab=0$, then $(2a+b)^2=9b^2$, so $2a+b=\pm 3b$. 
If $2a+b=3b$, then $a=b$, so $(a,b,c)=(k,k,0)$.
If $2a+b=-3b$, then $a=-2b$, so $(a,b,c)=\left(-2k,k,3k^2\right)$.
If $2a^2-b^2-ab=0$, then $b^2-2a^2+ab=0$ (symmetrical to the other case), so analogously $(a,b,c)=(k,k,0),\left(k,-2k,-3k^2\right)$.
All solutions are $(a,b,c)=(k,k,0),(k,-k,0),(-2k,k,3k^2),(k,-2k,-3k^2)$.
