Let A ={x:x³+1=0};B={x:x²-x+1=0} .Find A n (intersection) B when (i) x is real (ii) x is not real.

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{A}=\left\{\mathrm{x}:{\mathrm{x}}^3+1=0\right\} \\ \mathrm{B}=\left\{\mathrm{x}:{\mathrm{x}}^2-\mathrm{x}+1=0\right\} \\ \mathrm{Intersection} \mathrm{will} \mathrm{consist} \mathrm{of} \mathrm{common} \mathrm{point}, \mathrm{hence} \mathrm{to} \mathrm{get} \mathrm{intersection} \mathrm{of} \mathrm{two} \mathrm{sets}, \mathrm{put}: \\ {\mathrm{x}}^3+1={\mathrm{x}}^2-\mathrm{x}+1 \\ \Rightarrow \left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)={\mathrm{x}}^2-\mathrm{x}+1 \\ \Rightarrow \left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)-\left({\mathrm{x}}^2-\mathrm{x}+1\right)=0 \\ \Rightarrow \left({\mathrm{x}}^2-\mathrm{x}+1\right)\left(\mathrm{x}+1-1\right)=0 \\ \Rightarrow \left({\mathrm{x}}^2-\mathrm{x}+1\right)\left(\mathrm{x}\right)=0 \\ \mathbf{case}\left(\mathbf{i}\right)\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{real} \\ \mathrm{Discriminant} \mathrm{of} \left({\mathrm{x}}^2-\mathrm{x}+1\right) \mathrm{is} \mathrm{negative} \left[{\left(1-\right)}^2-4=-3\right], \mathrm{hence} \mathrm{it} \mathrm{cannot} \mathrm{be} 0. \\ \Rightarrow \mathrm{x}=0 \\ \mathrm{Hence} \\ \mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{=}\left\{\mathbf{ }\mathbf{0}\mathbf{ }\right\} \\ \left({\mathrm{x}}^2-\mathrm{x}+1\right)\left(\mathrm{x}\right)=0 \\ \mathbf{Case}\left(\mathbf{ii}\right)\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{not}\mathbf{ }\mathbf{real}\mathbf{.} \\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+1=0 \\ \mathrm{By} \mathrm{Shridharacharya}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{have}: \\ \mathrm{x}=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(1\right)}}{2} \\ =\frac{1\pm \sqrt{1-4}}{2} \\ =\frac{1\pm \sqrt{-3}}{2} \\ =\frac{1\pm \mathrm{i}\sqrt{3}}{2} \\ \Rightarrow \mathrm{x}=\frac{1+\mathrm{i}\sqrt{3}}{2} \mathrm{or} \frac{1-\mathrm{i}\sqrt{3}}{2} \\ \Rightarrow \mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{=}\left\{\frac{\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right\} \end{gathered}$$

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