\(\left(x^3-4\right)^3=\left(\sqrt[3]{\left(x^2+4\right)^2}+4\right)^2\)

\(pt\Leftrightarrow x^9-12x^6+48x^3-64=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+8\sqrt[3]{\left(x^2+4\right)^2}+16\)

\(\Leftrightarrow x^9-12x^6+48x^3-128=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2-16+8\sqrt[3]{\left(x^2+4\right)^2}-32\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\dfrac{\left(x^2+4\right)^4-4096}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\dfrac{512\left(x^2+4\right)^2-32768}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\dfrac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\right]=0\)

Dễ thấy: \(\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\dfrac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}-\dfrac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

P.s: dễ thấy thật :v