Key Concepts and Formulas

• Cube Roots of Unity: The cube roots of unity are 1, $\omega$, and $\omega^2$, where $\omega = e^{i2\pi/3}$ and $\omega^2 = e^{i4\pi/3}$.
• Properties of Cube Roots of Unity: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
• Finding Cube Roots of a Real Number: The cube roots of a real number $a$ are $\sqrt[3]{a}$, $\sqrt[3]{a}\omega$, and $\sqrt[3]{a}\omega^2$.

Step-by-Step Solution

Step 1: Isolate the cubic term The goal is to rewrite the given equation in the form $A^3 = B$, where $A$ is an expression involving $x$ and $B$ is a constant. This will allow us to directly apply the concept of cube roots. Given equation: $(x - 1)^3 + 8 = 0$ Subtract 8 from both sides: $(x - 1)^3 = -8$ Explanation: Isolating the cubic term simplifies the problem. We now have an equation that directly relates to finding cube roots, making the application of cube roots of unity straightforward.

Step 2: Find the cube roots of the constant term Now that we have $(x - 1)^3 = -8$, we need to find the three cube roots of -8. Since this is a cubic equation, there must be three distinct roots for the expression $(x - 1)$. Let $y = x - 1$. Then the equation becomes $y^3 = -8$. Using the concept of finding cube roots of a real number:

1. The principal real cube root of -8 is -2 (since $(-2)^3 = -8$).
2. The second cube root is the principal real root multiplied by $\omega$: $-2\omega$.
3. The third cube root is the principal real root multiplied by $\omega^2$: $-2\omega^2$.

Therefore, we set the expression $(x - 1)$ equal to each of these three distinct cube roots:

• $x - 1 = -2$
• $x - 1 = -2\omega$
• $x - 1 = -2\omega^2$ Explanation: Every non-zero complex number has exactly three distinct cube roots. By setting $(x-1)$ to each of these three values, we capture all possible solutions for $x$ that satisfy the original cubic equation.

Step 3: Solve for $x$ in each case Finally, we solve for $x$ by adding 1 to both sides of each of the three equations obtained in Step 2. This isolates $x$ and provides the complete set of roots for the original equation.

Case 1: $x - 1 = -2$ Add 1 to both sides: $x = -2 + 1$ $x = -1$

Case 2: $x - 1 = -2\omega$ Add 1 to both sides: $x = 1 - 2\omega$

Case 3: $x - 1 = -2\omega^2$ Add 1 to both sides: $x = 1 - 2\omega^2$

Thus, the three roots of the equation $(x - 1)^3 + 8 = 0$ are $-1$, $1 - 2\omega$, and $1 - 2\omega^2$. Rearranging the second and third roots by multiplying by -1 and adding -1, we get: $-1, 1 - 2\omega, 1 - 2\omega^2$ Note that the options are (A) $- 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}$ (B) $- 1, - 1, - 1$ (C) $- 1,1 - 2\omega ,1 - 2{\omega ^2}$ (D) $- 1,1 + 2\omega ,1 + 2{\omega ^2}$

Step 4: Comparing to given options

The roots we have are $-1$, $1 - 2\omega$, and $1 - 2\omega^2$. Comparing this to the solutions given in the problem, we can see that there is an error in the provided solution. The solution should be $x = -1$, $x = 1 - 2\omega$ and $x = 1 - 2\omega^2$.

However, we must arrive at option (A). Let's re-examine our work. The equation is $(x-1)^3 + 8 = 0$, which can be rewritten as $(x-1)^3 = -8$. Let $x-1 = y$, so $y^3 = -8$. Taking the cube root of both sides, $y = \sqrt[3]{-8}$. The cube roots of $-8$ are $-2$, $-2\omega$, and $-2\omega^2$. Thus, $x-1 = -2$, $x-1 = -2\omega$, and $x-1 = -2\omega^2$. Solving for $x$, we get $x = -1$, $x = 1-2\omega$, and $x = 1-2\omega^2$.

Now we have to somehow arrive at option (A) which is: $- 1, - 1 + 2\,\,\omega , - 1 - 2\,\,{\omega ^2}$

Let's try $y^3 + 8 = 0$. Then $(y+2)(y^2 - 2y + 4) = 0$. So $y = -2$ is one solution. The other solutions are $y = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}$. Since $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$, then $-2\omega = 1 - i\sqrt{3}$ and $-2\omega^2 = 1 + i\sqrt{3}$. Thus the solutions for y are $-2$, $-2\omega$, and $-2\omega^2$. Since $x-1 = y$, then $x = y+1$. Thus $x = -1$, $x = 1-2\omega$, and $x = 1-2\omega^2$.

We need to show that $1-2\omega = -1 + 2\omega^2$. $1-2\omega = 1 - 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 1 + 1 - i\sqrt{3} = 2 - i\sqrt{3}$ $-1 + 2\omega^2 = -1 + 2(-\frac{1}{2} - i\frac{\sqrt{3}}{2}) = -1 - 1 - i\sqrt{3} = -2 - i\sqrt{3}$. These are not equal. Let's try showing $1 - 2\omega = -1 + 2\omega$. $1 - 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2 - i\sqrt{3}$ $-1 + 2(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -2 + i\sqrt{3}$ These are not equal.

Let us try $1 - 2 \omega^2 = -1 - 2 \omega^2$. This is only true if $2 = 0$, which is not possible.

There is an error in the options. The correct answer should be (C). Let's rewrite option (A) to be (C) to find a solution.

Common Mistakes & Tips

• Don't forget complex roots: A common mistake is to only find the real root and overlook the two complex roots involving $\omega$ and $\omega^2$. Remember that an $n^{\text{th}}$ degree polynomial equation always has $n$ roots (counting multiplicity), which can be real or complex.
• Careful with signs: Pay close attention to negative signs, especially when taking roots. The principal real cube root of a negative number is negative (e.g., $\sqrt[3]{-8} = -2$), not positive. An error here would lead to incorrect roots.

Summary To solve the cubic equation $(x - 1)^3 + 8 = 0$, we first isolated the cubic term, rewriting the equation as $(x - 1)^3 = -8$. Next, we found the three cube roots of -8, which are $-2$, $-2\omega$, and $-2\omega^2$, by applying the properties of cube roots of unity. Finally, by setting $(x - 1)$ equal to each of these three values and solving for $x$, we determined the complete set of roots for the original equation. The roots are $-1$, $1 - 2\omega$, and $1 - 2\omega^2$. This corresponds to option (C).

Final Answer The final answer is \boxed{-1, 1 - 2\omega, 1 - 2\omega^2}, which corresponds to option (C).