<s>[INST] 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 $48+20y^2−6y^3$
 [/INST] $$x+3y=4 \implies x=4-3y \implies (4-3y)^2+3y^2=8 \implies 3y^2-6y+2=0 \implies y=1\pm \frac{1}{\sqrt{3}}$$
We get $x=1-\sqrt{3}$ for $y=1+\frac{1}{\sqrt{3}}$, then $F=x^3+3y^3=(1-\sqrt{3})^3+3(1+\frac{1}{\sqrt{3}})^3=16-\frac{8}{\sqrt{3}}.$
or $x=1+\sqrt{3}$ for $y=1-\frac{1}{\sqrt{3}}$
Finally, $F=x^3+3y^3=(1+\sqrt{3})^3+3(1-\frac{1}{\sqrt{3}})^3=16+\frac{8}{\sqrt{3}}$
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