Let $\chi$ be a character of $\mathbb F_p^\times$ of order $3$ that sends $a_0$ to $e^{ 2\pi i/3}$ . Then $\sum_{x\in \mathbb F_p} \chi(x) = D_0 + e^{ 2\pi i/3} D_1 + e^{-2\pi i/3} D_2$  which combined with $D_0+ D_1 + D_2 =p$ lets one calculate $D_0, D_1, D_2$ from $\sum_{x\in \mathbb F_p} \chi(x)$.

This sum $\sum_{x\in \mathbb F_p} \chi(x)$ may be expressed directly in terms of a cubic Gauss sum over $\mathbb F_{p^3}$: One takes the cubic Gauss sum, divides by $p$, and subtracts $1$. So this reduces the problem to computing Gauss sums, on which much literature exists.

Indeed, let $\alpha \in \mathbb F_{p^3}$ have minimal polynomial $f$. Let $N$ be the norm from $\mathbb F_{p^3}$ to $\mathbb F_p$. Then $f(x) = N(x-\alpha)$ so 

$$ \sum_{x\in \mathbb F_p} \chi(x)= \sum_{x\in \mathbb F_p} \chi(N(x-\alpha)) = \frac{1}{p-1} \sum_{x\in \mathbb F_p} \sum_{ y \in \mathbb F_{p}^\times}  \chi( y^3 N(x-\alpha)) = \frac{1}{p-1} \sum_{x\in \mathbb F_p} \sum_{ y \in \mathbb F_{p}^\times}  \chi(  N(xy-y\alpha)) = \frac{1}{p-1} \sum_{x\in \mathbb F_p} \sum_{ y \in \mathbb F_{p}^\times}  \chi(  N(x-y\alpha)) =\frac{1}{p-1} \sum_{x,y \in \mathbb F_p} \chi(  N(x-y\alpha))  -1. $$

Now let $\psi$ be an additive character of $\mathbb F_{p^3}$, for example $x\mapsto e^{ 2\pi i\operatorname{tr} x/p}$. Let $z \in \mathbb F_{p^3}^\times$ have the propery that $\psi( z \alpha ) =\psi(z)=1$. Then 

$$\frac{1}{p-1} \sum_{x,y \in \mathbb F_p} \chi(  N(x-y\alpha)) =\frac{1}{p(p-1)} \sum_{t \in \mathbb F_p} \sum_{a \in \mathbb F_{p^3}} \chi( N(a)) \psi( zt a) = \frac{1}{p(p-1)} \sum_{t \in \mathbb F_p^\times } \sum_{a \in \mathbb F_{p^3}} \chi( N(a)) \psi( zt a) =  \frac{1}{p(p-1)} \sum_{t \in \mathbb F_p^\times } \sum_{a \in \mathbb F_{p^3}} \chi( N(t^{-1} a)) \psi( z a) = \frac{1}{p(p-1)} \sum_{t \in \mathbb F_p^\times } \sum_{a \in \mathbb F_{p^3}} \chi( N( a)) \psi( z a) = \frac{1}{ p} \sum_{a \in \mathbb F_{p^3}} \chi( N( a)) \psi( z a)$$

which is a Gauss sum over $\mathbb F_{p^3}$.