If `x/y = y/z = z/w ` then `(x^m+y^m +z^m+w^m)/(x^(-m)+y^(-m )+z^(-m)+w^(-m))` is equals to -

Options

A. 0
B. 1
C. `(xyzw)^m`
D. `(xyzw)^(m//2)`

Answer

To solve the problem, we start with the given condition:

1. **Given**: \( \frac{x}{y} = \frac{y}{z} = \frac{z}{w} \)

Let's denote this common ratio as \( k \). Therefore, we can express \( x, y, z, \) and \( w \) in terms of \( k \):

\[ x = ky, \quad y = kz, \quad z = kw \]

From \( y = kz \), we can express \( z \) in terms of \( w \):

\[ y = kw \quad \Rightarrow \quad z = \frac{y}{k} = \frac{kw}{k} = w \]

Now substituting back, we have:

\[ z = w, \quad y = kw, \quad x = k^2w \]

2. **Substituting values**: Now we can substitute these expressions into the original equation we need to evaluate:

\[ \frac{x^m + y^m + z^m + w^m}{x^{-m} + y^{-m} + z^{-m} + w^{-m}} \]

Substituting for \( x, y, z, \) and \( w \):

\[ = \frac{(k^2w)^m + (kw)^m + w^m + w^m}{(k^2w)^{-m} + (kw)^{-m} + w^{-m} + w^{-m}} \]

Simplifying the numerator:

\[ = \frac{k^{2m}w^m + k^mw^m + w^m + w^m}{k^{-2m}w^{-m} + k^{-m}w^{-m} + w^{-m} + w^{-m}} \]

\[ = \frac{(k^{2m} + k^m + 2)w^m}{(k^{-2m} + k^{-m} + 2)w^{-m}} \]

3. **Simplifying further**: We can factor out \( w^m \) from the numerator and \( w^{-m} \) from the denominator:

\[ = \frac{(k^{2m} + k^m + 2)}{(k^{-2m} + k^{-m} + 2)} \cdot w^{m - (-m)} \]

\[ = w^{2m} \cdot \frac{(k^{2m} + k^m + 2)}{(k^{-2m} + k^{-m} + 2)} \]

4. **Finding the ratio**: Now, we need to simplify the fraction:

\[ = w^{2m} \cdot \frac{(k^{2m} + k^m + 2)}{\frac{1}{k^{2m}} + \frac{1}{k^m} + 2} \]

Multiplying the numerator and denominator by \( k^{2m} \):

\[ = w^{2m} \cdot \frac{k^{2m}(k^{2m} + k^m + 2)}{1 + k^m + 2k^{2m}} \]

5. **Final simplification**: Notice that the expression simplifies to a constant value:

\[ = w^{2m} \cdot k^{2m} \cdot \frac{k^{2m} + k^m + 2}{1 + k^m + 2k^{2m}} \]

Since \( k^{2m} + k^m + 2 \) and \( 1 + k^m + 2k^{2m} \) are symmetric in terms of \( k \), the overall expression simplifies to \( -1 \) when evaluated.

Thus, the final answer is:

\[ \frac{x^m + y^m + z^m + w^m}{x^{-m} + y^{-m} + z^{-m} + w^{-m}} = -1 \]