Key Concepts and Formulas

• Polynomial Roots: Finding roots by factoring or using the rational root theorem.
• Complex Cube Roots of Unity: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, where $\omega = \frac{-1 + i\sqrt{3}}{2}$.
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Identify a Root by Inspection We are given the cubic equation $x^3 - 2x^2 + 2x - 1 = 0$. We can try to find an integer root by testing factors of the constant term, which is -1. Possible integer roots are +1 and -1. Let $P(x) = x^3 - 2x^2 + 2x - 1$.

If $x = 1$, then $P(1) = (1)^3 - 2(1)^2 + 2(1) - 1 = 1 - 2 + 2 - 1 = 0$. Therefore, $x = 1$ is a root of the equation. Why: Finding a root by inspection simplifies the problem by allowing us to factor the polynomial.

Step 2: Factor the Polynomial Since $x = 1$ is a root, then $(x - 1)$ is a factor of the polynomial. We can perform polynomial division or synthetic division to find the other factor. $\frac{x^3 - 2x^2 + 2x - 1}{x-1} = x^2 - x + 1$ Thus, the equation can be factored as $(x - 1)(x^2 - x + 1) = 0$. Why: Factoring the polynomial allows us to find all the roots by solving the resulting factors.

Step 3: Solve the Quadratic Equation Now we solve the quadratic equation $x^2 - x + 1 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}$ So, the two complex roots are $x = \frac{1 + i\sqrt{3}}{2}$ and $x = \frac{1 - i\sqrt{3}}{2}$. Why: The quadratic formula provides the exact solutions for any quadratic equation.

Step 4: Express Roots in Terms of $\omega$ Recall that $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$. Also, $-\omega = \frac{1 - i\sqrt{3}}{2}$ and $-\omega^2 = \frac{1 + i\sqrt{3}}{2}$. Thus, the roots of the equation are $1$, $-\omega^2$, and $-\omega$. Why: Expressing the roots in terms of $\omega$ simplifies calculations involving their powers, as we can use the properties of $\omega$.

Step 5: Calculate the 162nd Power of Each Root We need to find the sum of the 162nd powers of the roots: $1^{162} + (-\omega^2)^{162} + (-\omega)^{162}$.

• $1^{162} = 1$
• $(-\omega^2)^{162} = (-1)^{162}(\omega^2)^{162} = \omega^{324} = (\omega^3)^{108} = 1^{108} = 1$
• $(-\omega)^{162} = (-1)^{162}(\omega)^{162} = \omega^{162} = (\omega^3)^{54} = 1^{54} = 1$ Why: Using the property $\omega^3 = 1$, we can simplify any power of $\omega$ by finding the remainder when the exponent is divided by 3. In this case, both 324 and 162 are divisible by 3, so $\omega^{324} = \omega^{162} = 1$.

Step 6: Sum the Powers The sum of the 162nd powers of the roots is: $1^{162} + (-\omega^2)^{162} + (-\omega)^{162} = 1 + 1 + 1 = 3$ Why: We sum the results obtained for each root's power to get the final answer.

Common Mistakes & Tips

• Tip: Look for simple integer roots first.
• Tip: Remember the properties of $\omega$: $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
• Common Mistake: Be careful with negative signs. $(-\omega)^{162} = \omega^{162}$ because 162 is even.

Summary We found the roots of the cubic equation by inspection and factoring. Then, we expressed the complex roots in terms of $\omega$, the complex cube root of unity. Finally, we calculated the 162nd power of each root and summed them to find the answer. The sum of the 162th power of the roots is 3.

Final Answer The final answer is \boxed{3}.