How can I prove this:
Prove that, if the polynomial $p(x)=ax^3+bx^2+cx+d$ has roots $x_1,x_2,x_3$, then $d=ax_1 x_2 x_3$.
As many have said in the comments ...
If a polynomial $p$ has a root $x_0$ then we can factor out that root, i.e. $p(x) = q(x) (x-x_0)$ where $q$ is a polynomial of degree one less than $p$.
This allows us to rewrite $p(x) = ax^3 + bx^2 + cx + d$ as $p(x) = k(x-x_1)(x-x_2)(x-x_3)$ where $k$ is some constant. Multiplying through, we see that the coefficient of the $x^3$ term is $k$, so $k = a$ ; and the constant term is $-kx_1x_2x_3$, so we see that $d = -ax_1x_2x_3$ , as desired.