If ` alpha, beta, gamma ` are the cube roots of ` p, p lt 0` then for any x, y and z
`(x alpha + y beta + z gamma)/( x beta + y gamma + z alpha)` =

Options

A. 1
B. `omega`
C. `omega^(2)`
D. none

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{x \alpha + y \beta + z \gamma}{x \beta + y \gamma + z \alpha} \]

where \(\alpha, \beta, \gamma\) are the cube roots of \(p\) and \(p < 0\).

### Step 1: Identify the cube roots of \(p\)

Given that \(p < 0\), we can express \(p\) in polar form. The cube roots of \(p\) can be expressed as:

\[ \alpha = p^{1/3} \text{cis} \left(0\right) = p^{1/3} \]
\[ \beta = p^{1/3} \text{cis} \left(\frac{2\pi}{3}\right) = p^{1/3} \omega \]
\[ \gamma = p^{1/3} \text{cis} \left(\frac{4\pi}{3}\right) = p^{1/3} \omega^2 \]

where \(\omega = e^{2\pi i / 3}\) is the primitive cube root of unity.

### Step 2: Substitute the values into the expression

Now we substitute \(\alpha\), \(\beta\), and \(\gamma\) into the expression:

\[ \frac{x \alpha + y \beta + z \gamma}{x \beta + y \gamma + z \alpha} = \frac{x p^{1/3} + y (p^{1/3} \omega) + z (p^{1/3} \omega^2)}{x (p^{1/3} \omega) + y (p^{1/3} \omega^2) + z p^{1/3}} \]

### Step 3: Factor out \(p^{1/3}\)

We can factor \(p^{1/3}\) out of both the numerator and denominator:

\[ = \frac{p^{1/3} \left(x + y \omega + z \omega^2\right)}{p^{1/3} \left(x \omega + y \omega^2 + z\right)} \]

This simplifies to:

\[ = \frac{x + y \omega + z \omega^2}{x \omega + y \omega^2 + z} \]

### Step 4: Use properties of cube roots of unity

Recall that \(1 + \omega + \omega^2 = 0\). We can multiply the numerator and denominator by \(\omega^2\):

\[ = \frac{\omega^2 (x + y \omega + z \omega^2)}{\omega^2 (x \omega + y \omega^2 + z)} = \frac{\omega^2 x + \omega^3 y + \omega^4 z}{\omega^3 x + \omega^4 y + \omega^2 z} \]

Since \(\omega^3 = 1\) and \(\omega^4 = \omega\), we have:

\[ = \frac{\omega^2 x + y + \omega z}{x + \omega y + \omega^2 z} \]

### Step 5: Simplify the expression

Notice that the numerator and denominator are now equal:

\[ = \frac{\omega^2 x + y + \omega z}{x + \omega y + \omega^2 z} = \omega^2 \]

Thus, the final result is:

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

### Final Answer

\(\omega^2\)