If a, b, c are rational and no two of them are equal, then the equations
`(b-c)x^(2)+(c-a)x+(a-b)=0`
and, `a(b-c)x^(2)+ b(c-a)x+c(a-b)=0`

Options

A. have rational roots and exactly one them is common
B. will be such that at least one has rational roots
C. have at least one root common.
D. no common root

Explanation

To solve the problem, we need to analyze the two given quadratic equations and determine the relationship between their roots.

### Step 1: Write down the equations
The two equations given are:
1. \((b - c)x^2 + (c - a)x + (a - b) = 0\)
2. \(a(b - c)x^2 + b(c - a)x + c(a - b) = 0\)

### Step 2: Find the sum of coefficients for both equations
For the first equation, the sum of the coefficients is:
\[ (b - c) + (c - a) + (a - b) = 0 \]
For the second equation, the sum of the coefficients is:
\[ a(b - c) + b(c - a) + c(a - b) = 0 \]

### Step 3: Determine if \(x = 1\) is a root
Since the sum of the coefficients of both equations is zero, we can conclude that \(x = 1\) is a root of both equations.

### Step 4: Express the product of roots
Let’s denote the roots of the first equation as \(\alpha\) and \(1\). The product of the roots of the first equation can be expressed as:
\[ \alpha \cdot 1 = \frac{a - b}{b - c} \]
Thus, \(\alpha = \frac{a - b}{b - c}\).

For the second equation, let’s denote its roots as \(\beta\) and \(1\). The product of the roots of the second equation can be expressed as:
\[ \beta \cdot 1 = \frac{c(a - b)}{a(b - c)} \]
Thus, \(\beta = \frac{c(a - b)}{a(b - c)}\).

### Step 5: Compare \(\alpha\) and \(\beta\)
We have:
\[ \alpha = \frac{a - b}{b - c} \]
\[ \beta = \frac{c(a - b)}{a(b - c)} \]
To check if \(\alpha\) can equal \(\beta\), we can set:
\[ \frac{a - b}{b - c} = \frac{c(a - b)}{a(b - c)} \]
Cross-multiplying gives:
\[ (a - b) \cdot a(b - c) = (b - c) \cdot c(a - b) \]
Since \(a\), \(b\), and \(c\) are distinct, we can conclude that \(\alpha \neq \beta\).

### Step 6: Conclusion
Thus, we have established that both equations have rational roots, and they share exactly one common root, which is \(x = 1\).