If x = (-3+iroot3)/2 is a complex number, then the value of (x^2+3x^2)^2(x^2+3x+1) is (1) -9/8 (2) 6 (3) -18 (4) 36

Solution:

Given x = (-3+i√3)/2

= (-1+i√3)/2-(1/1) = ω-1 [since ω = (-1+i√3)/2]

(x2+3x2)2(x2+3x+1)

(x2+3x2)2(x2+3x+1)

= [(ω-1)2+3(ω-1)2][(ω-1)2+3(ω-1)+1]

= (ω2+ω-2)22+ω-1)

= (-1-2)2(-1-1) (since ω2 +ω = -1

= (-3)2(-2)

= -2×9

= -18

Hence option (3) is the answer.

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