The value of :
(a+bw+cw^​2)/(b+cw+aw^​2) +(a+bw+cw^​2)/(c+aw+w^​2) is,
1). 1
2). -1
3). 2
4). -2

where, w represents omega.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} 1,\mathrm{\omega } \mathrm{and} {\mathrm{\omega }}^2 \mathrm{are} \mathrm{roots} \mathrm{of} \mathrm{the} \mathrm{equation} {\mathrm{x}}^3-1=0 \mathrm{and} \mathrm{are} \mathrm{known} \mathrm{as} \mathrm{cube} \mathrm{roots} \mathrm{of} \mathrm{unity}. \\ \mathrm{Now}, \mathrm{we} \mathrm{know} \mathrm{that} , \\ \mathbf{1}\mathbf{+}\mathit{\omega }\mathbf{+}\mathbf{ }{\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{0} \\ {\mathit{\omega }}^{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{ } \\ {\mathit{\omega }}^{\mathbf{3}\mathbf{n}}\mathbf{=}\mathbf{1}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{where}\mathbf{ }\mathbf{\text{'}}\mathbf{n}\mathbf{\text{'}}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{any}\mathbf{ }\mathbf{integer}\mathbf{.}\mathbf{ } \\ {\omega }^4={\omega }^3.\omega \\ \Rightarrow {\omega }^4=\omega \\ \mathrm{There} \mathrm{is} \mathrm{a} \mathrm{printing} \mathrm{error} \mathrm{in} \mathrm{your} \mathrm{question}. \mathrm{In} \mathrm{the} \mathrm{denominator} \mathrm{of} \mathrm{second} \mathrm{term} \mathrm{it} \mathrm{should} \mathrm{be} {\mathrm{b\omega }}^2 \mathrm{rather} \mathrm{than} {\mathrm{\omega }}^2. \\ Let S=\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2} \\ \Rightarrow S=\frac{\omega }{\omega }\left(\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}\right)+\frac{{\omega }^2}{{\omega }^2}\left(\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}\right) \\ \Rightarrow S=\omega \left(\frac{a+b\omega +c{\omega }^2}{b\omega +c{\omega }^2+a{\omega }^3}\right)+{\omega }^2\left(\frac{a+b\omega +c{\omega }^2}{c{\omega }^2+a{\omega }^3+b{\omega }^4}\right) \\ \mathrm{Now} \mathrm{putting} {\mathrm{\omega }}^3=1 \mathrm{and} {\mathrm{\omega }}^4=\mathrm{\omega } , \mathrm{we} \mathrm{get}, \\ \Rightarrow S=\omega \left(\frac{a+b\omega +c{\omega }^2}{b\omega +c{\omega }^2+a}\right)+{\omega }^2\left(\frac{a+b\omega +c{\omega }^2}{c{\omega }^2+a+b\omega }\right) \\ \Rightarrow S=\omega \left(\frac{a+b\omega +c{\omega }^2}{a+b\omega +c{\omega }^2}\right)+{\omega }^2\left(\frac{a+b\omega +c{\omega }^2}{a+b\omega +c{\omega }^2}\right) \\ \Rightarrow S=\omega +{\omega }^2 \\ \mathrm{Now} 1+\omega +{\omega }^2=0 \\ \Rightarrow \omega +{\omega }^2=-1 \\ \Rightarrow S=-1 \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{2}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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