Suppose $$\begin{vmatrix}a & a^2 & a^3-1 \\b & b^2 & b^3-1 \\ c & c^2 & c^3-1 \end{vmatrix}=0$$, where a, b and c care distince real numbers. If a = 3, then the value of abc is ________________.

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

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

Taking -1 common from the second matrix of the third row.

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

Taking a,b,c common from row 1, row 2 and row 3 respectively from the first matrix.

$$abc\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix}a & a^2 & 1 \\b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix}=0$$

Now, we will switch column 3 of the second matrix to column 1. Since with each switch the sign changes, and for this we have to switch the columns twice, thus the sign will remain the same as before.

$$abc\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}-\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}=0$$

$$(abc-1)\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}=0$$

Performing row operations on matrix, R2 -> R2-R1, and R3 -> R3-R1.

$$(abc-1)\begin{vmatrix}1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}=0$$

$$\left(abc-1\right)\left[\left(b-a\right)\left(c^2-a^2\right)-\left(c-a\right)\left(b^2-a^2\right)\right]$$

$$\left(abc-1\right)\left(b-a\right)\left(c-a\right)\left(c+a-b-a\right)$$

$$\left(abc-1\right)\left(b-a\right)\left(c-a\right)\left(c-b\right)$$

Since a,b,c are distinct values, thus (b-a), (c-a), and (c-b) can never be 0, therefore, only (abc-1) can be = 0.

=> $$abc-1=0$$

=> $$abc=1$$