If (4 b²+1/(b²))=2, then (8 b³+1/(b³))= ? · ABC26GN1276

$$\begin{aligned} \left(4 b^{2}+\frac{1}{b^{2}}\right)=2 & \Rightarrow\left(2 b+\frac{1}{b}\right)^{2}-4=2 \\ & \Rightarrow\left(2 b+\frac{1}{b}\right)^{2}=6 \Rightarrow\left(2 b+\frac{1}{b}\right)=\sqrt{6} \\ & \Rightarrow\left(2 b+\frac{1}{b}\right)^{3}=(\sqrt{6})^{3}=6 \sqrt{6} \\ & \Rightarrow 8 b^{3}+\frac{1}{b^{3}}+3 \times 2 b \times \frac{1}{b} \cdot\left(2 b+\frac{1}{b}\right)=6 \sqrt{6} \\ & \Rightarrow\left(8 b^{3}+\frac{1}{b^{3}}\right)+6 \sqrt{6}=6 \sqrt{6} \\ & \Rightarrow\left(8 b^{3}+\frac{1}{b^{3}}\right)=0 \end{aligned}$$