The equations `(b-c)x+(c-a)y+(a-b)=0` and `(b^(3)-c^(3))x+(c^(3)-a^(3))y+a^(3)-b^(3)=0` will represent the same line if

Options

A. b=c
B. c=a
C. a=b
D. `a+b+c=0`

Answer

To determine the conditions under which the equations

1. \((b-c)x + (c-a)y + (a-b) = 0\)
2. \((b^3 - c^3)x + (c^3 - a^3)y + (a^3 - b^3) = 0\)

represent the same line, we can analyze the two equations step by step.

### Step 1: Rewrite the first equation
The first equation can be expressed in the standard form of a line:
\[ (b-c)x + (c-a)y + (a-b) = 0 \]

### Step 2: Factor the second equation
The second equation involves cubic terms. We can use the identity for the difference of cubes:
\[ b^3 - c^3 = (b-c)(b^2 + bc + c^2) \]
\[ c^3 - a^3 = (c-a)(c^2 + ca + a^2) \]
\[ a^3 - b^3 = (a-b)(a^2 + ab + b^2) \]
Thus, we can rewrite the second equation as:
\[ (b-c)(b^2 + bc + c^2)x + (c-a)(c^2 + ca + a^2)y + (a-b)(a^2 + ab + b^2) = 0 \]

### Step 3: Compare coefficients
For the two equations to represent the same line, their coefficients must be proportional. This means we can set up the following relationships based on the coefficients of \(x\), \(y\), and the constant term:

\[ \frac{(b-c)}{(b^3 - c^3)} = \frac{(c-a)}{(c^3 - a^3)} = \frac{(a-b)}{(a^3 - b^3)} \]

### Step 4: Simplify the conditions
From the first part, we have:
\[ \frac{(b-c)}{(b-c)(b^2 + bc + c^2)} = \frac{(c-a)}{(c-a)(c^2 + ca + a^2)} = \frac{(a-b)}{(a-b)(a^2 + ab + b^2)} \]
This simplifies to:
\[ 1 = \frac{(c-a)}{(c^2 + ca + a^2)} = \frac{(a-b)}{(a^2 + ab + b^2)} \]

### Step 5: Solve for conditions
From the above relationships, we can derive two conditions:
1. \(b - c = 0\) or \(c - a = 0\) or \(a - b = 0\)
2. \(a + b + c = 0\)

### Conclusion
The equations represent the same line if either \(b = c\) or \(a + b + c = 0\).