↵​​ Pls answer this
If ω is imaginary cube root of unity and (a + bω + cω²)²⁰¹⁵ = (a + bω² + cω) where a,b,c are unequal real numbers, then a² + b² + c² – ab – bc – ca is equal to- (1) 0 (2) (3) 1 (4) –1

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)}^{2015}=\left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)...\left(\mathrm{i}\right) \\ \mathrm{Take} \mathrm{conjugate} \mathrm{on} \mathrm{bot} \mathrm{sides}, \mathrm{we} \mathrm{know} \mathrm{that} \overline{\mathrm{\omega }}={\mathrm{\omega }}^2 \mathrm{and} \overline{{\mathrm{\omega }}^2}=\mathrm{\omega }, \mathrm{hence} \\ {\left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)}^{2015}=\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)....\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)\times \left(\mathrm{ii}\right) \\ {\left\{\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\times \left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)\right\}}^{2015}=\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\times \left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right) \\ \Rightarrow \frac{{\left\{\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\times \left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)\right\}}^{2015}}{\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\times \left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)}=1 \\ \Rightarrow {\left\{\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\times \left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)\right\}}^{2014}=1 \\ \Rightarrow {\left[{\mathrm{a}}^2+{\mathrm{ab\omega }}^2+\mathrm{ac\omega }+\mathrm{ab\omega }+{\mathrm{b}}^2{\mathrm{\omega }}^3+{\mathrm{bc\omega }}^2+{\mathrm{ac\omega }}^2+{\mathrm{bc\omega }}^4+{\mathrm{c}}^2{\mathrm{\omega }}^3\right]}^{2014}=1 \\ \mathrm{Use} {\mathrm{\omega }}^3=1 \\ \Rightarrow {\left[{\mathrm{a}}^2+{\mathrm{ab\omega }}^2+\mathrm{ac\omega }+\mathrm{ab\omega }+{\mathrm{b}}^2+{\mathrm{bc\omega }}^2+{\mathrm{ac\omega }}^2+\mathrm{bc\omega }+{\mathrm{c}}^2\right]}^{2014}=1 \\ \Rightarrow {\left[{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2+\mathrm{ab}\left(\mathrm{\omega }+{\mathrm{\omega }}^2\right)+\mathrm{bc}\left(\mathrm{\omega }+{\mathrm{\omega }}^2\right)+\mathrm{ca}\left(\mathrm{\omega }+{\mathrm{\omega }}^2\right)\right]}^{2014}=1 \\ \mathrm{Use} \left(\mathrm{\omega }+{\mathrm{\omega }}^2\right)=-1 \\ \Rightarrow {\left[{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}\right]}^{2014}=1 \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}=1 \mathrm{or} -1 \left(\mathrm{Answer}\right) \end{gathered}$$

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