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Complex Numbers: Root Equation Find Value

JEE Maths question with a full step-by-step solution.

Question
If α\alpha is a root of the equation x2−x+1=0x^{2} - x + 1 = 0, then find the value of
∣α333+α666+α999∣.\left|\alpha^{333} + \alpha^{666} + \alpha^{999}\right| .
Solution
Answer: 1
Step 1: Find a simple power relation for α\alpha. Multiply the equation by (x+1)(x+1), which is the missing factor of the sum-of-cubes identity:
(x+1)(x2−x+1)=x3+1.(x+1)\left(x^{2}-x+1\right) = x^{3}+1 .
 Step 2: Substitute x=αx = \alpha. Since α2−α+1=0\alpha^{2}-\alpha+1 = 0, the product is 00:
α3+1=0  ⟹  α3=−1.\alpha^{3} + 1 = 0 \implies \alpha^{3} = -1 .
 (This is legitimate: x=−1x = -1 is not itself a root, as 1+1+1=3≠01+1+1 = 3 \ne 0.) Step 3: Reduce each exponent using α3=−1\alpha^{3} = -1. Note 333333, 666666 and 999999 are all multiples of 33:
α333=(α3)111=(−1)111=−1,\alpha^{333} = \left(\alpha^{3}\right)^{111} = (-1)^{111} = -1 ,
α666=(α3)222=(−1)222=+1,\alpha^{666} = \left(\alpha^{3}\right)^{222} = (-1)^{222} = +1 ,
α999=(α3)333=(−1)333=−1.\alpha^{999} = \left(\alpha^{3}\right)^{333} = (-1)^{333} = -1 .
 Step 4: Add them.
α333+α666+α999=−1+1−1=−1.\alpha^{333} + \alpha^{666} + \alpha^{999} = -1 + 1 - 1 = -1 .
 Step 5: Take the modulus.
∣−1∣=1.\left|-1\right| = 1 .
 Step 6: Note this works for either root, since only α3=−1\alpha^{3} = -1 was used. Answer: 11 (i.e. 1.001.00).
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