posted before.

Can you give link, please?

Why can't people use the search function BEFORE posting problems? Searching for "ab(a -b) a^3 +b^3 +ab cube" and restricting the results to the number theory section, you'll easily find this, for example...

First you can easily prove Vp(a)=Vp(b) is impossible. Now assume Vp(a)>Vp(b) now Vp(LHS)=Vp(a)+2Vp(b) and Vp(RHS)=3Vp(b) or Vp(a)+Vp(b) which are both smaller than Vp(LHS) so Vp(a)=2Vp(b) done.

Let $a=dx$, $b=dy$ where $(x,y) = 1$. Then $d^3xy(x-y)$ divides $d^2(dx^3 + dy^3 + xy)$, so $dxy(x-y)$ divides $dx^3+dy^3+xy$. So $d$ divides $xy$, but also $x$ dividing $dy^3$ implies $x$ divides $d$; similarly $y$ divides $d$ and so (as $(x,y) = 1$) $xy$ divides $d$. Hence $d=xy$ and so $ab = d^2xy = (xy)^3$, done.

@above is $d$ a natural number?

PaperMath wrote: @above is $d$ a natural number? Yes, it is the greatest common divisor of $a$ and $b$.

How do you know that $a=dx$ and $b=dy$ is the only way that satisfies the equation?

@above every two positive integers have a greatest common divisor (well known) - if $d$ is such for $a$ and $b$, then define $x=a/d$, $y=b/d$ to obtain what is written.

Wait thanks I was being dumb and forgot $1$ could be a GCF