Q.

Let  w1 be a complex cube root of unity. If (4+5w+6w2)n2+2+(6+5w2+4w)n2+2+(5+6w+4w2)n2+2=0  and nN. And  n1,100  then number of values of  n is

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answer is 33.

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Detailed Solution

(4+5w+6w2)n2+2(1+wn2+2+w2n2+4)=0 n=3λ, n3λ+1, 3λ+2 n=3,6,9,12,.......99

So, number of values of n is 33

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