<p>The roots of the equation \((x-1)^3 + 8 = 0\) are:</p>
<p>\(-1, 1+2\omega, 1+2\omega^2\)</p>
<p>\(-1, 1-2\omega, 1-2\omega^2\)</p>
<p>\(-1, 1-2\omega, 1+2\omega^2\)</p>
<p>\(-1, -1+2\omega, -1+2\omega^2\)</p>
Step-by-Step Solution
Key Concept: Recognize (x-1)³ + 8 = 0 as a sum of cubes: (x-1)³ + 2³ = 0. Apply the factorization a³ + b³ = (a+b)(a²-ab+b²) to find all three roots including complex ones.
<p><strong>Step 1:</strong> Rewrite the equation as a sum of cubes: $(x-1)^3 + 2^3 = 0$</p><p><strong>Step 2:</strong> Apply the sum of cubes formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ with $a = (x-1)$ and $b = 2$:</p><p>$[(x-1) + 2][(x-1)^2 - 2(x-1) + 4] = 0$</p><p>$(x+1)[(x-1)^2 - 2(x-1) + 4] = 0$</p><p><strong>Step 3:</strong> From the first factor: $x + 1 = 0 \Rightarrow x = -1$ (real root)</p><p><strong>Step 4:</strong> For the quadratic factor, expand: $(x-1)^2 - 2(x-1) + 4 = x^2 - 2x + 1 - 2x + 2 + 4 = x^2 - 4x + 7$</p><p><strong>Step 5:</strong> Solve $x^2 - 4x + 7 = 0$ using the quadratic formula:</p><p>$x = \frac{4 \pm \sqrt{16-28}}{2} = \frac{4 \pm \sqrt{-12}}{2} = \frac{4 \pm 2i\sqrt{3}}{2} = 2 \pm i\sqrt{3}$</p><p><strong>The three roots are:</strong> $x = -1$ and $x = 2 \pm i\sqrt{3}$</p><p>∴ Answer: B</p>
Correct Answer: B