Lời giải:
Để cho gọn đặt \(\sqrt[3]{x+2016}=a\). PT trở thành:
\(\sqrt[3]{a^3-1}+a+\sqrt[3]{a^3+1}=0\)
\(\Leftrightarrow (\sqrt[3]{a^3-1}+1)+a+(\sqrt[3]{a^3+1}-1)=0\)
\(\Leftrightarrow \frac{a^3}{\sqrt[3]{(a^3-1)^2}-\sqrt[3]{a^3-1}+1}+a+\frac{a^3}{\sqrt[3]{(a^3+1)^2}+\sqrt[3]{a^3+1}+1}=0\)
\(\Leftrightarrow a( \frac{a^2}{\sqrt[3]{(a^3-1)^2}-\sqrt[3]{a^3-1}+1}+1+\frac{a^2}{\sqrt[3]{(a^3+1)^2}+\sqrt[3]{a^3+1}+1})=0\)
Ta thấy:
\(\sqrt[3]{(a^3-1)^2}-\sqrt[3]{a^3-1}+1=(\sqrt[3]{a^3-1}-\frac{1}{2})^2+\frac{3}{4}>0\)
\(\Rightarrow \frac{a^2}{\sqrt[3]{(a^3-1)^2}-\sqrt[3]{a^3-1}+1}\geq 0\)
Tương tự: \(\frac{a^2}{\sqrt[3]{(a^3+1)^2}+\sqrt[3]{a^3+1}+1}\geq 0\)
Do đó biểu thức " trong ngoặc " lớn hơn $0$
\(\Rightarrow a=0\)
\(\Rightarrow \sqrt[3]{x+2016}=0\Rightarrow x=-2016\)