Đặt \(\left\{{}\begin{matrix}\sqrt[3]{12-x}=a\\\sqrt[3]{4+x}=b\end{matrix}\right.\) ta có hệ:
\(\left\{{}\begin{matrix}a+b=2\\a^3+b^3=16\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=2\\\left(a+b\right)\left(a^2+b^2-ab\right)=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=2-a\\a^2+b^2-ab=8\end{matrix}\right.\)
\(\Rightarrow a^2+\left(2-a\right)^2-a\left(2-a\right)-8=0\)
\(\Leftrightarrow3a^2-6a-4=0\Rightarrow a=\frac{3\pm\sqrt{21}}{2}\)
\(\Rightarrow\sqrt[3]{12-x}=\frac{3\pm\sqrt{21}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{36-16\sqrt{21}}{9}\\x=\frac{36+16\sqrt{21}}{9}\end{matrix}\right.\)
Bài toán có tới 2 nghiệm thỏa mãn? b có 2 giá trị là \(\pm16\) lấy cái nào?
