May you solve this very interesting system of equations for me?
$d+a=-4$
$ad+b=0$
$bd+c=0$
$cd=3$
Simply replacing ends up in quartic equation, for example for $d^4+4d^3+3=0$. But I want a more intelligent solution!!!
May you please find for me a trick to reduce the system's equation into a cubic equation, for example $x^3+ax^2+bx+c=0$ where $x$ is a new variable related to the starting system? ( please not $d^3+3d^2-3d+3=0$).
This solution is the solution I seek for. Thank you a lot for your time!
It's not entirely clear what you want. However, it's fairly straightforward to eliminate variables in any way you like.
Eliminating $a$, then $b$, then $c$ gives your quartic in $d$, which factors as $$(d + 1) (d^2 + 3 d^2 - 2 d + 3) = 0 \tag{1}$$ ... but you don't want that. Well, eliminating $a$, $b$, $d$ gives $$(c + 3) (c^3 - 3 c^2 + 9 c + 9) = 0 \tag{2}$$ Better? If not, then $a$, $c$, $d$ gives $$(b + 3) (b^3 - 3 b^2 + 15 b + 3) = 0 \tag{3}$$ Or perhaps $d$, $b$, $c$, which yields $$(a+3) (a^3+9a^2+21 a + 1) = 0 \tag{4}$$
As it happens, $$a=b = c=-3 \qquad d = -1$$ is a solution to the system. Letting Mathematica solve the system, it looks like each of the roots of the cubics contributes to a solution; the exact expressions are messy, but the decimal forms are as follows: $$a = -0.0486\ldots \quad b = -0.1921\ldots \quad c = -0.7592\ldots \quad d = -3.9513\ldots \\[8pt] a = -4.4756\ldots \pm i\;0.7300\ldots \quad b = 1.5960\ldots \mp i\;3.6146\ldots \\ c = \phantom{-}1.8796\ldots \pm i\;2.8846\ldots \quad d = 0.4756\ldots \mp i\;0.7300\ldots $$