A question from a math olympiad book . Find the sum of the 5th powers of the roots of the equation $x^3+3x+9=0$
I tried using vieta relations and got that if the roots are a , b , c then
$a^4+b^4+c^4 = 36$. I dont know how to proceed forward .Please help

Here is a quite simple solution which is also much easier than the recommendations linked in the comment section:
For any solution $t \in \{a,b,c\}$ of $x^3 + 3x+9 = 0$ it follows:
$$t^3 = -3t-9 \Rightarrow t^5 = -3t^3-9t^2 \quad (1)$$
Now, note that
$$t(t^2+3)=-9 \Rightarrow t^2 = -\frac 9t -3 \quad (2)$$
$(1)$ and $(2)$ together gives
$$t^5 = -3(-3t-9)-9\left(-\frac 9t -3\right) = 9t + \frac{81}t+54 \quad (3)$$
Noting that $\frac 1t$ is a solution to $x^3+\frac 13 x^2 +\frac 19 = 0$, you finally obtain from $(3)$ using Vieta (sum of zeros):
$$a^5+b^5+c^5 = 9\cdot 0 + 81\cdot \left(-\frac 13\right) + 3\cdot 54 = 135$$
