If `bc+ca+ab=18` and
`|(1,a^(2),a^(3)),(1,b^(2),b^(3)),(1,c^(2),c^(3))|=lamda|(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))|`
the value of `lamda` is

Options

A. `-1`
B. 0
C. 9
D. 18

Answer

To solve the given problem step by step, we will start with the determinants provided and use properties of determinants to find the value of \(\lambda\).

### Step 1: Write down the determinants
We have two determinants given in the problem:

1. \( D_1 = \begin{vmatrix} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{vmatrix} \)
2. \( D_2 = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} \)

### Step 2: Evaluate \(D_1\)
The determinant \(D_1\) can be evaluated using the formula for determinants of the form:

\[ \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{vmatrix} = (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) \]

In our case, we have \(x_1 = a\), \(x_2 = b\), and \(x_3 = c\). Therefore:

\[ D_1 = (a - b)(b - c)(c - a) \]

### Step 3: Evaluate \(D_2\)
The determinant \(D_2\) can also be evaluated similarly:

\[ D_2 = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b - a)(c - a)(c - b) \]

### Step 4: Set up the equation
According to the problem, we have:

\[ D_1 = \lambda D_2 \]

Substituting the values we found for \(D_1\) and \(D_2\):

\[ (a - b)(b - c)(c - a) = \lambda (b - a)(c - a)(c - b) \]

### Step 5: Simplify the equation
Notice that we can factor out \((a - b)\), \((b - c)\), and \((c - a)\) from both sides. This gives us:

\[ 1 = \lambda \]

### Step 6: Use the condition \(bc + ca + ab = 18\)
From the problem, we know that \(bc + ca + ab = 18\). This condition does not directly affect our calculation of \(\lambda\) since we have already simplified the determinants.

### Final Result
Thus, the value of \(\lambda\) is:

\[ \lambda = 18 \]