If a,b,c are all unequal and different from 1 and the points `((t^(3))/(t-1),(t^(2)-3)/(t-1)),t=a,b,c` are collinear then `ab+bc+ca=`

Options

A. abc
B. `-abc`
C. `abc+3Sigma a`
D. None

Answer

To solve the problem, we need to determine the value of \( ab + bc + ca \) given that the points \(\left(\frac{t^3}{t-1}, \frac{t^2 - 3}{t-1}\right)\) for \(t = a, b, c\) are collinear.

### Step-by-Step Solution:

1. **Identify the Points**:
We have three points based on the values of \(t\):
- Point 1: \(\left(\frac{a^3}{a-1}, \frac{a^2 - 3}{a-1}\right)\)
- Point 2: \(\left(\frac{b^3}{b-1}, \frac{b^2 - 3}{b-1}\right)\)
- Point 3: \(\left(\frac{c^3}{c-1}, \frac{c^2 - 3}{c-1}\right)\)

2. **Collinearity Condition**:
The points are collinear if the area formed by them is zero. The area can be calculated using the determinant:
\[ \text{Area} = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = 0 \]
Substituting our points:
\[ \frac{1}{2} \begin{vmatrix} \frac{a^3}{a-1} & \frac{a^2 - 3}{a-1} & 1 \\ \frac{b^3}{b-1} & \frac{b^2 - 3}{b-1} & 1 \\ \frac{c^3}{c-1} & \frac{c^2 - 3}{c-1} & 1 \end{vmatrix} = 0 \]

3. **Multiply Rows by Denominators**:
To simplify, multiply the first row by \(a-1\), the second row by \(b-1\), and the third row by \(c-1\):
\[ \begin{vmatrix} a^3 & a^2 - 3 & a - 1 \\ b^3 & b^2 - 3 & b - 1 \\ c^3 & c^2 - 3 & c - 1 \end{vmatrix} = 0 \]

4. **Apply Row Operations**:
Perform row operations to simplify the determinant:
- R1 - R2
- R2 - R3

This results in:
\[ \begin{vmatrix} a^3 - b^3 & (a^2 - 3) - (b^2 - 3) & (a - 1) - (b - 1) \\ b^3 - c^3 & (b^2 - 3) - (c^2 - 3) & (b - 1) - (c - 1) \\ c^3 & c^2 - 3 & c - 1 \end{vmatrix} = 0 \]

5. **Factor the Determinant**:
The determinant can be factored using the identities for the differences of cubes and squares:
\[ (a-b)(a^2 + ab + b^2) \quad \text{and} \quad (b-c)(b^2 + bc + c^2) \]
This gives:
\[ (a-b)(b-c)(c-a) \cdot \text{(some polynomial in a, b, c)} = 0 \]

6. **Set the Polynomial to Zero**:
The polynomial must equal zero for the determinant to be zero. This leads us to the equation:
\[ ab + ac + bc - abc = a + b + c \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ ab + ac + bc = abc + (a + b + c) \]

8. **Conclusion**:
Thus, we find that:
\[ ab + ac + bc = abc + (a + b + c) \]

### Final Answer:
The value of \(ab + bc + ca\) is equal to \(abc + (a + b + c)\).