If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ is a perfect cube then consider $d>1$ then there is a $p$ that is prime and $p \mid d$.We have :
$p\mid a-b \Rightarrow p \mid (2b)^3 \Rightarrow p \mid 2b \Rightarrow p\mid 2$ or $p\mid b$
If $p\mid b$ then also $p\mid a$ (as $p\mid a-b$ holds). Then we will get to $p \mid 1$ because:
$ p \mid a^2+b^2+ab+a+b,p \mid a^2+b^2+ab+a+b+1 \Rightarrow p \mid 1$
which is clearly wrong then we have $p\mid 2$ so $p=2$ means $d=2^k$ where $k$ is a natural number including $0$.In the case $d=1$ we have the right result.So assemble $k \ge 1$.Because $2 \mid a-b$ we can conclude that $a,b$ have the same parity @Ghartal showed in his answer that if $a,b$ are both even we have a right result but if $a,b$ are both odd we don't.So maybe we have to prove $a,b$ can,t be both odd.

Revised proof:
We have $(a-b)(a^2 + b^2 + ab + a + b + 1) = (2b)^3$
From OP, we see that if $p$ divides $d$ then $p$ divides $2b \implies p \mid 2$ or $p \mid b$. 
Assume $p$ doesn't divide $2$. 
$p \mid b \implies p \mid a$ but then $p$ does not divide $a^2+b^2+ab +a+b+1$. 
Therefore $p=1$ or $p=2$. If $p=1$ then $a-b$ is a cube. 
If $p=2$ then since $p$ does not divide $b$ both $a$ and $b$ are odd. 
Now $a^2 + b^2 + ab + a + b + 1 \ne 0 \mod 4$. Therefore largest power
 of $2$ in $a^2 + b^2 + ab + a + b + 1$ is $2$. 
Largest power of $2$ in $a-b$ should be $4$ (as $b$ is odd)
So one possibility is $a-b = 4m^3$ and $a^2 + b^2 + ab + a + b + 1 = 2n^3$ where $b = mn$
I guess I am stuck at this point. 
