Simultaneous algebra question Let $x$, $y$ be real numbers such that
$$x+3y=4$$$$x^2+3y^2=8$$
Find $$x^3+3y^3$$
Also, let $$ax+by=v_1$$$$ax^2+by^2=v_2$$
Find the value of $ax^3+by^3$ in terms of $v_1$, $v_2$
I've tried $(x+3y)(x^2+9y^2−3xy)−6y^3$ and was left with $48+20y^2−6y^3$

Being lazy, I changed a little the notations
$$ax+by=v \tag 1$$ $$ax^2+by^2=w \tag 2$$
From $(1)$, you have $y=\frac{v-a x}{b}$. Plug in $(2)$ to face
$$\left(v^2-b w\right)-2 a vx+a\left(a+ b\right)x^2=0$$
$$x_\pm=\frac{av\pm\sqrt{a b \left(w (a+b)-v^2\right)}}{a (a+b)}\qquad \qquad y_\pm=\frac{v-a x_\pm}{b}$$
Now, just a bit of patience for the simplifications and
$$a x_+^3+b y_+^3=\frac{a^2 \left(bv+\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3+b^2 \left(a
   v-\sqrt{a b \left(w (a+b)-v^2\right)}\right)^3}{a^2 b^2 (a+b)^3}$$
