Let's integrate

Let, $f(x) =\displaystyle \int (3ax^2+2bx+c) \, dx=ax^3+bx^2+cx$ $\color{teal} {\text{[We won't need the integral constant as it's a definite integral.]}}$

Given that, $\displaystyle \int_{0}^3 (3ax^2+2bx+c) \, dx=\displaystyle \int_{1}^3 (3ax^2+2bx+c) \, dx$ $\Rightarrow \left[ax^3+bx^2+cx\right]_0^3=\left[ax^3+bx^2+cx\right ]_1^3$ $\Rightarrow f(3)-f(0)=f(3)-f(1)$ $\Rightarrow f(1)=f(0)$ $\Rightarrow \color{#0C6AC7} {\boxed {a+b+c=0}}$

Thinking about it logically, as in with the basic area formula length times width, we have two areas equalling each other that have different lengths. It's like saying 3x = 2x. The only solution for the width is zero. Therefore, a, b, and c have to be zero.

Then of course, 0 + 0 + 0 = 0

Let $f(x) = 3ax^2 + 2bx + c$

$\displaystyle{\int_0^3 f(x) \,dx} = \displaystyle{\int_0^1 f(x) \,dx} + \displaystyle{\int_1^3 f(x) \,dx} = \displaystyle{\int_1^3 f(x) \,dx}$ so that $0 = \displaystyle{\int_0^1 f(x) \,dx} = ax^3 + bx^2 + cx \Big|_0^1 = a+b+c$