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 value of abc +1 is

Answer

To solve the problem, we need to analyze the determinant and the conditions given in the question step by step.

### Step 1: Write the Determinant
We start with the determinant given in the problem:

\[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix} \]

We also have the additional rows:

\[ D' = \begin{vmatrix} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \\ \end{vmatrix} \]

### Step 2: Expand the Determinant
Using properties of determinants, we can manipulate the determinant \(D'\):

\[ D' = \begin{vmatrix} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \\ \end{vmatrix} \]

We can express \(1 + a^3\) as \(1 + a^3 = (1 + a)(1 + a^2 - a)\) and similarly for \(b\) and \(c\).

### Step 3: Factor Out Common Terms
We can factor out \(a\), \(b\), and \(c\) from the first column:

\[ D' = abc \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \\ \end{vmatrix} \]

### Step 4: Determine the Value of the Determinant
The determinant \(D\) is known to be non-zero since the vectors \((1, a, a^2)\), \((1, b, b^2)\), and \((1, c, c^2)\) are non-coplanar. Thus, we can conclude that:

\[ D \neq 0 \]

### Step 5: Set Up the Equation
Since we know that the determinant \(D'\) equals zero:

\[ abc \cdot D = 0 \]

Given that \(D \neq 0\), we must have:

\[ abc = -1 \]

### Step 6: Find the Value of \(abc + 1\)
Now we need to find the value of \(abc + 1\):

\[ abc + 1 = -1 + 1 = 0 \]

### Final Answer
Thus, the value of \(abc + 1\) is:

\[ \boxed{0} \]