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If ω(≠1) is cube root of unity satisfying 1a+ω+1b+ω+1c+ω=2ω2 and 1a+ω2+1b+ω2+1c+ω2=2ω, then the value of 1a+1+1b+1+1c+1 is : |
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Answer» If ω(≠1) is cube root of unity satisfying 1a+ω+1b+ω+1c+ω=2ω2 and 1a+ω2+1b+ω2+1c+ω2=2ω, then the value of 1a+1+1b+1+1c+1 is : |
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