Let ω and ω2 be the non-real cube roots of unity and 1/(a+ω)+1/(b+ω)+1/(c+ω)=2ω2 and 1/(a+ω2)+1/(b+ω2)+1/(c+ω2)=2ω, then calculate 1/(a+1)+1/(b+1)+1/(c+1).

Let ω and ω2 be the non-real cube roots of unity and 1/(a+ω)+1/(b+ω)+1/(c+ω)=2ω2 and 1/(a+ω2)+1/(b+ω2)+1/(c+ω2)=2ω, then calculate 1/(a+1)+1/(b+1)+1/(c+1). Correct Answer 2

Given relations can be written as: 1/(a+ω)+1/(b+ω)+1/(c+ω)=2/ω, and 1/(a+ω2)+1/(b+ω2)+1/(c+ω2)=2/ω2⇒ω and ω2 are roots of 1/(a+x)+1/(b+x)+1/(c+x)=2/x. ⇒/=2/x⇒x3+(bc+ca+ab)x-2abc=0. Let 3rd root be α (apart from ω and ω2). Then, α+ ω+ ω2=coefficient of x2=0⇒α=1. Hence, 1/(a+1)+1/(b+1)+1/(c+1)=2/1=2.

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