$a,b,c$ are distinct real roots of $x^3-3x+1=0$. $a^8+b^8+c^8$ is

$ \textbf{(A)}\ 156 \qquad \textbf{(B)}\ 171 \qquad \textbf{(C)}\ 180 \qquad \textbf{(D)}\ 186 \qquad \textbf{(E)}\ 201$
Given that \(a, b, c\) are distinct real roots of the polynomial equation \(x^3 - 3x + 1 = 0\), we need to find the value of \(a^8 + b^8 + c^8\).

1. **Identify the polynomial and its roots:**
   The polynomial is \(P(x) = x^3 - 3x + 1\). The roots of this polynomial are \(a, b, c\).

2. **Use the properties of polynomial roots:**
   By Vieta's formulas, for the polynomial \(x^3 - 3x + 1 = 0\), we have:
   \[
   a + b + c = 0,
   \]
   \[
   ab + bc + ca = -3,
   \]
   \[
   abc = -1.
   \]

3. **Transform the polynomial:**
   We need to find \(a^8 + b^8 + c^8\). To do this, we will use the fact that if \(r_i\) are roots of \(P(x) = x^3 - 3x + 1\), then \(r_i^2\) are roots of a related polynomial. Specifically, if \(u, v, w\) are roots of \(x^3 - 3x + 1\), then \(u^2, v^2, w^2\) are roots of a polynomial derived from \(P(x)\).

4. **Find the polynomial for \(u^2, v^2, w^2\):**
   Consider the polynomial \(Q(x) = x^3 - 6x^2 + 9x - 1\). This polynomial has roots \(u^2, v^2, w^2\).

5. **Find the polynomial for \(u^4, v^4, w^4\):**
   Similarly, \(u^4, v^4, w^4\) are roots of the polynomial \(R(x) = x^3 - 18x^2 + 69x - 1\).

6. **Find the polynomial for \(u^8, v^8, w^8\):**
   Finally, \(u^8, v^8, w^8\) are roots of the polynomial \(S(x) = x^3 - 186x^2 + \ldots\).

7. **Sum of the roots:**
   By Vieta's formulas, the sum of the roots of the polynomial \(S(x) = x^3 - 186x^2 + \ldots\) is given by the coefficient of \(x^2\) with a negative sign. Therefore, the sum of the roots \(u^8 + v^8 + w^8\) is:
   \[
   a^8 + b^8 + c^8 = 186.
   \]

The final answer is \(\boxed{186}\).