ĐKXĐ: \(x^8\le17\)
Đặt \(\left\{{}\begin{matrix}\sqrt[4]{17-x^8}=a\ge0\\\sqrt[3]{2x^8-1}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=1\\2a^4+b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=a-1\\2a^4+b^3-33=0\end{matrix}\right.\)
\(\Leftrightarrow2a^4+\left(a-1\right)^3-33=0\)
\(\Leftrightarrow2a^4+a^3-3a^2+3a-34=0\)
\(\Leftrightarrow\left(a-2\right)\left(2a^3+5a^2+7a+17\right)=0\)
\(\Leftrightarrow a=2\) (ngoặc sau luôn dương với \(a\ge0\))
\(\Leftrightarrow\sqrt[4]{17-x^8}=2\Leftrightarrow17-x^8=16\)
\(\Leftrightarrow x^8=1\Rightarrow x=\pm1\)
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