lf ω is a complex cube root of unity, then the value of a+bω+cω2c+aω+bω2+a+bω+cω2b+cω+aω2is

a+bw+cw2c+aw+bw2+a+bw+cw2b+cw+aw2 =w2(aw+bw2+c)c+aw+bw2+w(aw2+b+cw)b+cw+aw2 =w2+w =-1

lf ω is a complex cube root of unity, then the value of a+bω+cω2c+aω+bω2+a+bω+cω2b+cω+aω2is

a+bw+cw2c+aw+bw2+a+bw+cw2b+cw+aw2 =w2(aw+bw2+c)c+aw+bw2+w(aw2+b+cw)b+cw+aw2 =w2+w =-1

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$lf ω is a complex cube root of unity, then the value of \frac{a + bω + {cω}^{2}}{c + aω + {bω}^{2}} + \frac{a + bω + {cω}^{2}}{b + cω + {aω}^{2}} is$

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answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{a + bw + {cw}^{2}}{c + aw + {bw}^{2}} + \frac{a + bw + {cw}^{2}}{b + cw + {aw}^{2}} \\ = & \frac{w^{2} (aw + {bw}^{2} + c)}{c + aw + {bw}^{2}} + \frac{w ({aw}^{2} + b + cw)}{b + cw + {aw}^{2}} \\ = & w^{2} + w \\ = & - 1 \end{array}$