Solving degree 3 equations

Question

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?

Answer 1

Yes there is a simpler way, since these equations are homogeneous we can substitute $$y=ax$$ which will make the equation simpler to solve.

Substituting it in both the equations we get, $$x^3[4+3a+a^3]=8$$ and $$x^3[2-2a+a^2]=1$$,

Dividing these two and simplifying we get $$a^3-8a^2+19a-12=0$$

Which factors to $$[a-1][a-3][a-4]=0$$

Putting $a=1$ in either of the equations we get $$x=y=1$$

I believe you can now put $a=3,4$ in the equation yourself and get the answer

Answer 2

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: $$\left\{(1,1),\left(\frac{1}{\sqrt[3]5},\frac{3}{\sqrt[3]5}\right), \left(\frac{1}{\sqrt[3]{10}},\frac{4}{\sqrt[3]{10}}\right)\right\}$$