We start by considering the given equation: (z−100)3+1000=0 This can be rewritten as: (z−100)3=−1000 Noting that -1000 is the same as −103, we get: (z−100)3=(−10)3 We can now take the cube root of both sides: z−100=−10 z−100=−10ω z−100=−10ω2 Here, ω and ω2 are the other two primitive cube roots of unity. Solving these equations for z, we get: For z−100=−10 : z=90 For z−100=−10ω : z=100−10ω For z−100=−10ω2 : z=100−10ω2 To find the expressions in the given options, recall the properties of the cube roots of unity. If ω is a primitive cube root of unity, then we have: ω3=1 1+ω+ω2=0 This means we can express 1 as: 1=−ω−ω2 Substituting 1=−ω−ω2 in the solutions we found: For z=100−10ω : z=10(10−ω) For z=100−10ω2. z=10(10−ω2) So, the solutions are: 10(10−ω),10(10−ω2), and 90