Three roots in equation

First solving it for general.

$K + L + M = \alpha$

$KL + LM + KM = \beta$

$KLM = \gamma$

$(K + L + M)^2 = K^2 + L^2 + M^2 + 2(KL + LM + KM)$

$\Rightarrow K^2 + L^2 + M^2 = \alpha^2 - 2\beta\hspace{15pt} (= \alpha_1)$

$(KL + LM + KM)^2 = K^2L^2 + L^2M^2 + K^2M^2 + 2KLM(K + L + M)$

$\Rightarrow K^2L^2 + L^2M^2 + K^2M^2 = \beta^2 - 2\alpha\gamma \hspace{15pt}(= \beta_1)$

$K^2L^2M^2 = (KLM)^2 = \gamma^2 \hspace{15pt}(= \gamma_1)$

So, now we have

$K^2 + L^2 + M^2 = \alpha_1$

$K^2L^2 + L^2M^2 + K^2M^2 = \beta_1$

$K^2L^2M^2 = \gamma_1$

Doing the same thing again and again we can get

$K^{2^n} + L^{2^n} + M^{2^n}$ for any $n \geq 0\;,\,n\in\mathbb{Z}$

Taking $\alpha = \Large\frac{-9}{4}$

$\beta = \Large\frac{-5}{4}$

$\gamma = -3$

Doing the same procedure for $3$ times we can get

$K^8 + L^8 + M^8 = \large\frac{127096129}{256^2}$ $\approx 1939.333$

The problem can be solved using Newton's sums method. Since the roots of $4x^3+5x^2-9x+12=0$ are $K$ , $L$ , $M$ , and $d$ . Then by Vieta's formula , $S_1 = K+L+M = - \frac 54$ , $S_2 = KL+LM+MK = -\frac 94$ , $S_3 = KLM = - 3$ . Let $P_n = K^n+L^n+M^n$ , where $n$ is a positive integer and we need to find $P_8$ . Then we have:

$\begin{array} {ll} P_1 = S_1 & = -\dfrac 54 \\ P_2 = S_1P_1 - 2S_2 & = \dfrac {97}{16} \\ P_3 = S_1P_2 - S_2P_1 + 3S_3 & = -\dfrac {1241}{64} \\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 & = \dfrac {10657}{256} \\ P_5 = S_1P_4 - S_2P_3 + S_3P_2 & = -\dfrac {116585}{1024} \\ P_6 = S_1P_5 - S_2P_4 + S_3P_3 & = \dfrac {1204849}{4096} \\ P_7 = S_1P_6 - S_2P_5 + S_3P_4 & = -\dfrac {12267449}{16384} \\ P_8 = S_1P_7 - S_2P_6 + S_3P_5 & = \dfrac {127096129}{65536} \end{array}$

Therefore $P_8 = \dfrac {127096129}{65536} \approx \boxed{1939.333}$ . We can do the computation easily with an Excel spreadsheet.