number theory - Prime factor of $A=14^7+14^2+1$

Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.

Answer

Hint: I've seen the 3rd cyclotomic polynomial too many times.

$$\begin{aligned} x^7+x^2+1&=(x^7-x^4)+(x^4+x^2+1)\\ &=x^4(x^3-1)+\frac{x^6-1}{x^2-1}\\ &=x^4(x+1)(x^2+x+1)+\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}\\ &=x^4(x+1)(x^2+x+1)+(x^2+x+1)(x^2-x+1) \end{aligned}$$