If `(x)/(2a-3b)=(y)/(3b-4c)=(z)/(4c-2a)`, then find the value of `x^(3)+y^(3)+z^(3)`.

Answer

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

\[ \frac{x}{2a - 3b} = \frac{y}{3b - 4c} = \frac{z}{4c - 2a} = k \]

where \( k \) is a constant.

From this, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

1. \( x = k(2a - 3b) \)
2. \( y = k(3b - 4c) \)
3. \( z = k(4c - 2a) \)

Next, we will find \( x + y + z \):

\[ x + y + z = k(2a - 3b) + k(3b - 4c) + k(4c - 2a) \]

Combining the terms inside the parentheses:

\[ x + y + z = k \left( (2a - 2a) + (-3b + 3b) + (-4c + 4c) \right) \]

This simplifies to:

\[ x + y + z = k(0) = 0 \]

Since \( x + y + z = 0 \), we can use the identity for the sum of cubes:

\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]

Given that \( x + y + z = 0 \), we have:

\[ x^3 + y^3 + z^3 = 3xyz \]

Now, we need to find \( xyz \):

\[ xyz = k(2a - 3b) \cdot k(3b - 4c) \cdot k(4c - 2a) = k^3(2a - 3b)(3b - 4c)(4c - 2a) \]

Thus, we can express \( x^3 + y^3 + z^3 \):

\[ x^3 + y^3 + z^3 = 3k^3(2a - 3b)(3b - 4c)(4c - 2a) \]

This is the final expression for \( x^3 + y^3 + z^3 \).

### Final Answer:
\[ x^3 + y^3 + z^3 = 3k^3(2a - 3b)(3b - 4c)(4c - 2a) \]