Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers.
I thought about proving that this number is divisible by $2$, $3$ and $7$ but I don't know how should I do that.

Put $E=(a^6+b^6-1)(a^6+b^6-2)$; the two factors are consecutive so coprime and $252=2^2\cdot3^2\cdot7$.
$1)$ $E\equiv0\pmod7$. By Fermat's Little Theorem it is quite clear,  even if $a$ or $b$ is divisible by $7$.
$2)$ $E\equiv0\pmod{3^2}$. We know that $$(X\pm1)^6=X^6\pm6X^5+15X^4\pm20X^3+15X^2\pm6X+1$$ There are two cases to consider:
$► (a,b)=(3m,3n\pm1)\Rightarrow(3m)^6+(3n\pm1)^6-1\equiv\pm6(3n)\equiv0\pmod{3^2}$.
$► (a,b)=(3m\pm1,3n\pm1)\Rightarrow(3m\pm1)^6+(3n\pm1)^6-2\equiv\pm6(3m)\pm6(3n)\equiv0\pmod{3^2}$.
$3) E\equiv0\pmod{2^2}$. There are two cases to consider:
$► (a,b)=(2m,2n+1)\Rightarrow(2m)^6+(2n+1)^6-1\equiv6(2n)\equiv0\pmod4.$
$►(a,b)=(2m+1,2n+1)\Rightarrow(2m+1)^6+(2n+1)^6-2\equiv6(2m+2n)\equiv0\pmod4$.
(In both, $2)$ and $3)$,we have used one of the two factors of $E$ to prove the divisibility).
