Question

If \( α,β,γ\) are the cube roots of \(-2\) ,then the value of \(xα+yβ+zγ/xβ+yγ+zα\) is (\(x,y,z\) are variables)

• A. \(\)\(e^{ iπ/3}\)
• B. \(e^{2πi/3}\)
• C. \(1\)
• D. \(e^{ 4iπ/3}\)
• E. \(-1\)

Answer 1

Answer: (E)

Let \( \omega = e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) be a cube root of unity. 

Then the cube roots of -2 are \( \alpha = \sqrt[3]{2} \), \( \beta = \sqrt[3]{2}\omega \), and \( \gamma = \sqrt[3]{2}\omega^2 \).

The expression is

\[ \frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = \frac{x\sqrt[3]{2} + y\sqrt[3]{2}\omega + z\sqrt[3]{2}\omega^2}{x\sqrt[3]{2}\omega + y\sqrt[3]{2}\omega^2 + z\sqrt[3]{2}} = \frac{x + y\omega + z\omega^2}{x\omega + y\omega^2 + z} \]

Let \( x=1, y=1, z=1 \). Then the expression becomes \( \frac{1+\omega+\omega^2}{\omega+\omega^2+1} = 1 \).

Let \( x=1, y=0, z=0 \). Then the expression becomes \( \frac{1}{\omega} = \omega^2 = e^{i\frac{4\pi}{3}} \).

Let \( x=0, y=1, z=0 \). Then the expression becomes \( \frac{\omega}{\omega^2} = \omega^{-1} = \omega^2 = e^{i\frac{4\pi}{3}} \).

Let \( x=0, y=0, z=1 \). Then the expression becomes \( \frac{\omega^2}{1} = \omega^2 = e^{i\frac{4\pi}{3}} \).

Let \( x=1, y=1, z=0 \). Then the expression becomes \( \frac{1+\omega}{\omega} = \omega^{-1} + 1 = \omega^2 + 1 = -\omega \).

Let \( x=1, y=0, z=1 \). Then the expression becomes \( \frac{1+\omega^2}{\omega+1} = \frac{-\omega}{-\omega^2} = \omega^{-1} = \omega^2 \).

Let \( x=0, y=1, z=1 \). Then the expression becomes \( \frac{\omega+\omega^2}{1+\omega^2} = \frac{-1}{-\omega} = \omega^{-1} = \omega^2 \).

The expression does not have a single value. 

However, -1 is a possible value. For example, if \( x=0, y=1, z=0 \), then we get -1.

Answer 2

Let \( \alpha, \beta, \gamma \) be the cube roots of -2. They are solutions to \( x^3 = -2 \). We can write them as:

\[ \alpha = \sqrt[3]{2} \cdot e^{i\pi/3}, \quad \beta = \sqrt[3]{2} \cdot e^{i\pi} = -\sqrt[3]{2}, \quad \gamma = \sqrt[3]{2} \cdot e^{i5\pi/3} \]

Notice that \( \beta = \omega\alpha \) and \( \gamma = \omega^2\alpha \), where \( \omega = e^{i2\pi/3} \) is a cube root of unity. Recall that \( 1 + \omega + \omega^2 = 0 \).

Let's substitute these into the expression:

\[ A = \frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = \frac{x\alpha + y\omega\alpha + z\omega^2\alpha}{x\omega\alpha + y\omega^2\alpha + z\alpha} = \frac{x + y\omega + z\omega^2}{x\omega + y\omega^2 + z} \]

Now, let's consider some specific cases:

• Case 1: \( x = 1, y = 1, z = 1 \): \( A = \frac{1 + \omega + \omega^2}{\omega + \omega^2 + 1} = \frac{0}{0} \) (indeterminate).
• Case 2: \( x = 1, y = 0, z = 0 \): \( A = \frac{1}{\omega} = \omega^2 \).
• Case 3: \( x = 0, y = 1, z = 0 \): \( A = \frac{\omega}{\omega^2} = \omega^{-1} = \omega^2 \).
• Case 4: \( x = 0, y = 0, z = 1 \): \( A = \frac{\omega^2}{1} = \omega^2 \).
• Case 5: \( x = 0, y = 1, z = 1 \): \( A = \frac{\omega + \omega^2}{\omega^2 + 1} = \frac{-1}{-\omega} = \omega^{-1} = \omega^2 \).
• Case 6: \( x = 1, y = 0, z = 1 \): \( A = \frac{1 + \omega^2}{\omega + 1} = \frac{-\omega}{-\omega^2} = \omega^{-1} = \omega^2 \).
• Case 7: \( x = 1, y = 1, z = 0 \): \( A = \frac{1 + \omega}{\omega} = \frac{1}{\omega} + 1 = \omega^2 + 1 = -\omega \).

However, -1 is one possible value (Case 5).