If `a,b,c in R` such that `a^3+b^3+c^3=2` and `|[a,b,c],[b,c,a],[c,a,b]|=0` then find `abc`

Options

A. `2/3`
B. `-2/3`
C. `1/3`
D. `1`

Answer

To solve the problem, we need to find the value of \( abc \) given the conditions \( a^3 + b^3 + c^3 = 2 \) and the determinant \( |[a,b,c],[b,c,a],[c,a,b]| = 0 \).

### Step-by-Step Solution:

1. **Understanding the Determinant Condition**:
The determinant condition \( |[a,b,c],[b,c,a],[c,a,b]| = 0 \) implies that the rows (or columns) of the matrix are linearly dependent. This means that there exist constants \( x, y, z \) (not all zero) such that:
\[ x[a, b, c] + y[b, c, a] + z[c, a, b] = 0 \]
This can also be interpreted as a relation among \( a, b, c \).

2. **Calculating the Determinant**:
We can compute the determinant:
\[ |[a,b,c],[b,c,a],[c,a,b]| = a(bc - ac) - b(bc - ab) + c(ab - b^2) \]
Expanding this gives:
\[ = abc - a^2b - b^2c + ab^2 + ac^2 - b^2c = 0 \]
Rearranging terms leads to:
\[ a^2b + b^2c + c^2a = abc \]

3. **Using the Identity for Cubes**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Given \( a^3 + b^3 + c^3 = 2 \), we have:
\[ 2 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

4. **Finding Relationships**:
Since \( a^2b + b^2c + c^2a = abc \), we can substitute \( abc \) in the identity:
\[ 2 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
This means \( a+b+c \) and \( a^2 + b^2 + c^2 - ab - ac - bc \) must also satisfy certain conditions.

5. **Expressing \( abc \)**:
From the determinant condition and the cube sum, we can derive:
\[ abc = \frac{a^3 + b^3 + c^3}{3} \]
Substituting the known value:
\[ abc = \frac{2}{3} \]

### Final Result:
Thus, the value of \( abc \) is:
\[ \boxed{\frac{2}{3}} \]