Suppose `a,b,c, gt0` and
`|(a^(3)-1,a^(2),a),(b^(3)-1,b^(2),b),(c^(3)-1,c^(2),c)|=0` then least possible value of `a+b+c` is ____________

Answer

To solve the problem, we need to analyze the determinant given in the question:

\[ \begin{vmatrix} a^3 - 1 & a^2 & a \\ b^3 - 1 & b^2 & b \\ c^3 - 1 & c^2 & c \end{vmatrix} = 0 \]

### Step 1: Rewrite the determinant

We can rewrite the first column of the determinant as follows:

\[ a^3 - 1 = (a - 1)(a^2 + a + 1) \]

Thus, we can express the determinant as:

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

### Step 2: Factor out common terms

Since we are interested in the determinant being zero, we can factor out \( (a - 1), (b - 1), (c - 1) \) from the first column:

\[ (a - 1)(b - 1)(c - 1) \begin{vmatrix} a^2 + a + 1 & a^2 & a \\ b^2 + b + 1 & b^2 & b \\ c^2 + c + 1 & c^2 & c \end{vmatrix} = 0 \]

### Step 3: Analyze the determinant condition

For the determinant to be zero, either \( a - 1 = 0 \), \( b - 1 = 0 \), or \( c - 1 = 0 \). This means at least one of \( a, b, c \) must be equal to 1.

### Step 4: Use AM-GM inequality

Assuming \( a = 1 \) (without loss of generality), we want to minimize \( a + b + c \). By the AM-GM inequality:

\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]

Given that \( a = 1 \), we have:

\[ \frac{1 + b + c}{3} \geq \sqrt[3]{1 \cdot b \cdot c} = \sqrt[3]{bc} \]

### Step 5: Set \( b \) and \( c \) to their minimum values

To minimize \( 1 + b + c \), we can set \( b = 1 \) and \( c = 1 \):

\[ 1 + 1 + 1 = 3 \]

### Conclusion

Thus, the least possible value of \( a + b + c \) is:

\[ \boxed{3} \]