If `t_(1),t_(2) and t_(3)` are distinct, the points `(t_(1)2at_(1)+at_(1)^(3)), (t_(2),2"at"_(2)+"at_(2)^(3)) and (t_(3) ,2at_(3)+at_(3)^(3))`

Options

A. `t_(1)t_(2)t_(3)=1`
B. `t_(1)+t_(2)+t_(3)=t_(1)t_(2)t_(3)`
C. `t_(1)+t_(2)+t_(3)=0`
D. `t_(1)+t_(2)+t_(3)=-1`

Explanation

To solve the problem, we need to check the collinearity of the points given by the coordinates \((t_1, 2at_1 + at_1^3)\), \((t_2, 2at_2 + at_2^3)\), and \((t_3, 2at_3 + at_3^3)\). The condition for collinearity of three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is that the determinant of the matrix formed by these points must be equal to zero.

### Step-by-Step Solution:

1. **Set up the determinant**:
We need to form the determinant using the coordinates of the three points:
\[ \begin{vmatrix} t_1 & 2at_1 + at_1^3 & 1 \\ t_2 & 2at_2 + at_2^3 & 1 \\ t_3 & 2at_3 + at_3^3 & 1 \end{vmatrix} = 0 \]

2. **Calculate the determinant**:
The determinant can be expanded as follows:
\[ = t_1 \begin{vmatrix} 2at_2 + at_2^3 & 1 \\ 2at_3 + at_3^3 & 1 \end{vmatrix} - t_2 \begin{vmatrix} 2at_1 + at_1^3 & 1 \\ 2at_3 + at_3^3 & 1 \end{vmatrix} + t_3 \begin{vmatrix} 2at_1 + at_1^3 & 1 \\ 2at_2 + at_2^3 & 1 \end{vmatrix} \]

3. **Evaluate the 2x2 determinants**:
Each of the 2x2 determinants can be calculated:
\[ \begin{vmatrix} 2at_2 + at_2^3 & 1 \\ 2at_3 + at_3^3 & 1 \end{vmatrix} = (2at_2 + at_2^3) - (2at_3 + at_3^3) = 2a(t_2 - t_3) + a(t_2^3 - t_3^3) \]
Using the identity \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\), we can rewrite \(t_2^3 - t_3^3\):
\[ = (t_2 - t_3)(t_2^2 + t_2t_3 + t_3^2) \]
Thus,
\[ = 2a(t_2 - t_3) + a(t_2 - t_3)(t_2^2 + t_2t_3 + t_3^2) = (t_2 - t_3)(2a + a(t_2^2 + t_2t_3 + t_3^2)) \]

4. **Substituting back into the determinant**:
Substitute back into the determinant:
\[ t_1(t_2 - t_3)(2a + a(t_2^2 + t_2t_3 + t_3^2)) - t_2(t_1 - t_3)(2a + a(t_1^2 + t_1t_3 + t_3^2)) + t_3(t_1 - t_2)(2a + a(t_1^2 + t_1t_2 + t_2^2)) = 0 \]

5. **Factor out common terms**:
Since \(t_1\), \(t_2\), and \(t_3\) are distinct, we can factor out the differences:
\[ (t_2 - t_1)(t_3 - t_1)(t_3 - t_2)(\text{some expression}) = 0 \]
This implies that the remaining expression must equal zero.

6. **Conclusion**:
After simplification, we find that the condition for collinearity leads us to:
\[ t_1 + t_2 + t_3 = 0 \]

### Final Answer:
Thus, the condition for the points to be collinear is:
\[ t_1 + t_2 + t_3 = 0 \]