Let $p(x),q(x)$ be distinct polynomials with real coefficients such that the sum of the coefficients of both polynomials equals $S$. If $ (p(x))^3-(q(x))^3=p(x^3)-q(x^3)$, then prove the following:
(a) $p(x)-q(x)=(x-1)^ar(x)$ for some integer $a ≥ 1$ and a polynomial $r(x)$ with $r(1)\ne0$.
(b) $S^2=3^{a-1}$
Since $p(1)=q(1)=S$, $x=1$ is a root of $p(x)-q(x)$. Hence, one can write $$p(x)-q(x)=(x-1)^a r(x),\ a\ge 1$$ where $r(x)$ is a polynomial with $r(1)\ne 0$. Now, from the given condition we can write, using this form of $p(x)-q(x)$ $$(x-1)^a r(x)(p^2(x)+q^2(x)+p(x)q(x))=(x^3-1)^a r(x^3)\\\Rightarrow r(x)(p^2(x)+q^2(x)+p(x)q(x))=(x^2+x+1)^ar(x^3)$$ Putting $x=1$ in both the sides, and cancelling $r(1)$ since $r(1)\ne 0$ $$(p^2(1)+q^2(1)+p(1)q(1))=3^a\Rightarrow 3S^2=3^a$$ The answer follows.
