<s>[INST] Solving degree 3 equations Solve for $x$,
\begin{cases}4x^3+3x^2y+y^3=8\\
2x^3-2x^2y+xy^2=1\end{cases}
I tried substitution of $x$, but it got very complex.
Is there a simpler way to do this?
 [/INST] We obtain:
$$4x^3+3x^2y+y^3=8(2x^3-2x^2y+xy^2)$$ or
$$12x^3-19x^2y+8xy^2-y^3=0$$ or
$$12x^3-12x^2y-7x^2y+7xy^2+xy^2-y^3=0$$ or
$$(x-y)(12x^2-7xy+y^2)=0$$ or
$$(x-y)(3x-y)(4x-y)=0.$$
Can you end it now?
I got the following answer:
$$\left\{(1,1),\left(\frac{1}{\sqrt[3]5},\frac{3}{\sqrt[3]5}\right), \left(\frac{1}{\sqrt[3]{10}},\frac{4}{\sqrt[3]{10}}\right)\right\}$$
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