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Class-9»Maths

Please answer the question with steps.

Solution : w = \frac{- 1 + \sqrt{3} i}{2}; w^{2} = \frac{- 1 - \sqrt{3} i}{2} {[\frac{- 1 + \sqrt{3} i}{2}]}^{- 8} + {[\frac{- 1 - \sqrt{3} i}{2}]}^{- 8} = w^{- 8} + {(w^{2})}^{- 8} we know that, w^{3} = 1 and {(a^{m})}^{n} = a^{mn} = w^{- 8} + w^{- 16} = \frac{1}{w^{8}} + \frac{1}{w^{16}} = \frac{w}{w^{8} \times w} + \frac{w^{2}}{w^{16} \times w^{2}} we know that, a^{m} \times a^{n} = a^{m + n} = \frac{w}{w^{9}} + \frac{w^{2}}{w^{18}} = \frac{w}{{(w^{3})}^{3}} + \frac{w^{2}}{{(w^{3})}^{5}} = w + w^{2} We know that, w^{2} + w + 1 = 0 So, w^{2} + w = - 1 Option B

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