Find the six complex roots of the equation $f(x)=x^6+8\sqrt{2}x^3+64=0$

Question

My work for this $$f(x) = x^6+8\sqrt{2}x^3+64=0$$ is $$ (x^3)^2+8\sqrt{2}(x^3)+64=0 \\x^3=\frac{-8\sqrt{2}\pm\sqrt{128-4(64)}}{2(1)} \\x^3=-4\sqrt{2}\pm4\sqrt{2}i$$ $$\text{Let }\ x=r(\cos\theta+i\sin\theta), \\|x^3|=|x|^3=\sqrt{32+32}=\sqrt{64}=8 \rightarrow |x|=r=2 \\\text{When }\ x^3=-4\sqrt{2}+4\sqrt{2}i, \\\text{By using De Moivre's Theorem}, \\2^3(\cos\theta+i\sin\theta)^3=-4\sqrt{2}+4\sqrt{2}i \\8(\cos3\theta+i\sin3\theta)=-4\sqrt{2}+4\sqrt{2}i \\\cos3\theta+i\sin3\theta=\frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i \\\therefore \cos3\theta=\frac{-1}{\sqrt{2}}\ \text{ and }\ \sin3\theta=\frac{1}{\sqrt{2}} \\\text{Principle value of }\ 3\theta=\frac{3\pi}{4},\frac{11\pi}{4},\frac{19\pi}{4} \\\theta=\frac{\pi}{4},\frac{11\pi}{12},\frac{19\pi}{12} $$ $$ \text{Since coefficients of }\ f(x)\ \text{are all real,}\ x\ \text{is a complex root}, \\\therefore \overline{x}\ \text{is also the other root by Conjugate Root Theorem}, $$ $$\therefore \text{The six sixth roots are:} \\x_{1}=2(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\sqrt{2}+\sqrt{2}i) \\\text{so on},\ x_{2}=-\frac{\sqrt{3}+1}{\sqrt{2}}+\frac{\sqrt{3}-1}{\sqrt{2}}i \\x_{3}=\frac{\sqrt{3}-1}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}i \\x_{4}=\overline{x_1}=\sqrt{2}-\sqrt{2}i \\x_{5}=\overline{x_2}=-\frac{\sqrt{3}+1}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}i \\x_6=\overline{x_3}=\frac{\sqrt{3}-1}{\sqrt{2}}-\frac{\sqrt{3}+1}{\sqrt{2}}i $$ Somehow these answers are wrong. The correct answers are: $$ \frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i,\ -\frac{\sqrt{3}+1}{2\sqrt{2}}\pm\frac{\sqrt{3}-1}{2\sqrt{2}}i,\ \frac{\sqrt{3}-1}{2\sqrt{2}}\pm\frac{\sqrt{3}+1}{2\sqrt{2}}i $$ What have I done wrong? Can anyone point it out, please?