Let $f(x)=(x+a)(x+b)(x+c) +(x-a)(x-b)(x-c)$. Find all factors of $f(x)$ Let $f(x)=(x+a)(x+b)(x+c) +(x-a)(x-b)(x-c)$. Find all factors of $f(x)$. Hence find the remainder of $2020 \times 2025\times 2027$ when divided by 2017.
When $x=0$ , $f(0)=0$. Therefore $(x-0)$ is a factor. 
Then $f(x)\equiv x(Ax^2+Bx+C) \equiv (x+a)(x+b)(x+c) +(x-a)(x-b)(x-c)$ 
Thus by comparing coefficients we have that $$A=2, B=0 , C=ab+bc+ac$$
How can I find the remainder by putting suitable values for $x,a,b,c$ ?

HINT: prove the $f(x)$ has the form $$f(x)=2\,x \left( ab+ac+bc+{x}^{2} \right) $$
