System of Polynomial Equations

Given that

$\begin{aligned} x^3-3x^2y-6xy^2 + 8y^3 & = 20 & \small \blue{\text{Also given that }x^2 + xy-2y^2 = 4} \\ (x+2y)^3 - 9x^2y - 18xy^2 & = 5(x^2+xy-2y^2) \\ (x+2y)^3 - 9xy(x+2y) & = 5(x-y)(x+2y) & \small \blue{\text{For }x+2y \ne 0} \\ (x+2y)^2 - 9xy & = 5(x-y) \\ x^2 - 5xy + 4y^2 & = 5(x-y) \\ (x-y)(x-4y) & = 5(x-y) & \small \blue{\text{for }x-y \ne 0} \\ x-4y & = 5 \\ \implies x & = 4y+5 \end{aligned}$

Substituting in $x^2 + xy - 2y^2 = 4$ :

$\begin{aligned} (6y+5)(3y+5) & = 4 \\ 18y^2 + 45y - 21 & = 0 \\ 6y^2 + 15y - 7 & = 0 \end{aligned}$

By Vieta's formula , the sum of roots $Y = -\frac {15}6 = - \frac 52$ . Then the sum of roots of $x$ , $X = 5 + 4y_1 + 5+4y_2 = 10+4Y = 10+4\left(-\frac 52\right) = 0$ . And $5X + 4Y = 5(0) + 4 \left(-\frac 52\right) = \boxed{-10}$ . (Note that $x+2y \ne 0$ and $x-y \ne 0$ .)

Let $A=x^3-3x^2y-6xy^2+8y^3=20$ and $B=x^2+xy-2y^2=4$ . Notice how the $2y^2$ and $x^2$ terms in Equation $B$ divide the $8y^3$ and $x^3$ terms in Equation $A$ , respectively, with a remainder of $0$ . As a result, performing Polynomial Division would be a good idea. Dividing $A$ by $B$ by Long Polynomial Division (or Factoring both equations) yields $C=x-4y=5$ and substituting this back into equation $B$ yields the sum of the possible values of $y$ as $-2.5$ by Vieta's formula, so $Y=-2.5$ . Using equation $C$ , the sum of all possible values of $x$ is $0$ , so $X=0$ . Thus, $5X+4Y=0-10=-10$ which is the final answer.