Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+5}=a\\\sqrt[3]{x+6}=b\end{matrix}\right.\)
\(pt\Leftrightarrow a+b=\sqrt[3]{a^3+b^3}\)
\(\Leftrightarrow\left(a+b\right)^3=a^3+b^3\)
\(\Leftrightarrow\left(a+b\right)\left(a^2+2ab+b^2-a^2-b^2+ab\right)=0\)
\(\Leftrightarrow\left(a+b\right).3ab=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\a=0\\b=0\end{matrix}\right.\)
TH1: \(a=-b\)
\(\Leftrightarrow\sqrt[3]{x+5}=-\sqrt[3]{x+6}\)
\(\Leftrightarrow x+5=-x-6\)
\(\Leftrightarrow x=-\frac{11}{2}\)
TH2: \(a=0\Leftrightarrow\sqrt[3]{x+5}=0\Leftrightarrow x=-5\)
TH3: \(b=0\Leftrightarrow\sqrt[3]{x+6}=0\Leftrightarrow x=-6\)
Vậy phương trình có 3 nghiệm ...
