Prove that $108^3-7^3$ is a multiple of $101$

Question

Encountered the next problem: prove that $108^3-7^3$ is the multiple to $101$.

As I understand if $c$ is the multiple to $a$ it means that there exist $b\in\mathbb{N}$ that $ab=c$. So i want to factor out $101$ from the expression, and I don't see a way to do it. I will appreciate any ideas.

Answer 1

Hint: $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 2

Use the properties of congruences, so you don't have to do any factorization.

Since $108\equiv 7\pmod{101}$, you have $$ 108^3\equiv7^3\pmod{101} $$ which is the same as saying that $108^3-7^3$ is a multiple of $101$.