If `|(a,a^2,1+a^3),(b,b^2,1+b^3),(c,c^2,1+c^3)|=0` and vectors `(1,a,a^2),(1,b,b^2) and (1,c,c^2)` are non coplanar then the product abc equals (A) 2 (B) -1 (C) 1 (D) 0

Answer

To solve the given problem, we need to analyze the determinant and the conditions provided. Let's break it down step by step.

### Step 1: Write the Determinant
The determinant given in the problem is:
\[ D = \begin{vmatrix} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{vmatrix} \]
We know that \( |D| = 0 \).

### Step 2: Rewrite the Determinant
We can rewrite the third column of the determinant:
\[ 1 + a^3 = 1 + a \cdot a^2 \]
Thus, we can express the determinant as:
\[ D = \begin{vmatrix} a & a^2 & 1 + a \cdot a^2 \\ b & b^2 & 1 + b \cdot b^2 \\ c & c^2 & 1 + c \cdot c^2 \end{vmatrix} \]

### Step 3: Apply Properties of Determinants
We can separate the determinant into two parts:
\[ D = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a \\ b & b^2 & b \\ c & c^2 & c \end{vmatrix} \]
The second determinant can be factored out:
\[ = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + abc \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]
Since the second determinant is zero, we have:
\[ D = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = 0 \]

### Step 4: Non-Coplanarity Condition
The vectors \( (1, a, a^2) \), \( (1, b, b^2) \), and \( (1, c, c^2) \) are given to be non-coplanar. This means that the determinant of these vectors must not equal zero:
\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \neq 0 \]

### Step 5: Conclusion from the Determinant
Since we have \( D = 0 \) and the non-coplanarity condition implies that the determinant of the vectors must not be zero, we can conclude that:
\[ 1 + abc = 0 \implies abc = -1 \]

### Final Answer
Thus, the product \( abc \) equals \( -1 \). The correct option is (B) -1.

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