\(\Leftrightarrow\int^{x^3-y^3-3\left(x-y\right)=0}_{x^6+y^6=1}\Leftrightarrow\int^{\left(x-y\right)\left(x^2+xy+y^2-3\right)=0\left(\cdot\right)}_{x^6+y^6=1\left(\cdot\cdot\right)}\)
(*) <=> x =y hoặc x2 +xy +y2 =3
+ Nếu x =y thì (**) <=> => 2 x6 =1 => x6 =1/2=>x =\(\sqrt[6]{\frac{1}{2}}\);hoặc x =-\(\sqrt[6]{\frac{1}{2}}\)
+Nếu x2 +xy +y2 =3 =>\(\int^{\left(x^2+y^2\right)+xy=3}_{x^6+y^6=1}\Leftrightarrow\int^{x^2+y^2+xy=3}_{\left(x^2+y^2\right)\left(\left(x^2+y^2\right)^2-3x^2y^2\right)=1}\Leftrightarrow\int^{s+p=3}_{s\left(s^2-3p^2\right)=1\left(\cdot\cdot\cdot\right)}\)
=> p = 3 -s (***) => s( s2 - 3 ( s2-6s +9)) =1=>s(-2s2+18s -27) =1 => 2s3 -18s2 +27s +1 =0 => Nghiệm lẻ thế
