If a, b, c are distinct real numbers and \(\begin{vmatrix}a & a^{2} & a^{3} - 1\\ b & b^{2} & b^{3} - 1\\ c & c^{2} & c^{3} - 1\end{vmatrix} = 0\), then
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(abc = 1\)
The third column is a difference, so the determinant splits into two:
\( \displaystyle \Delta = \begin{vmatrix}a & a^{2} & a^{3}\\ b & b^{2} & b^{3}\\ c & c^{2} & c^{3}\end{vmatrix} - \begin{vmatrix}a & a^{2} & 1\\ b & b^{2} & 1\\ c & c^{2} & 1\end{vmatrix} \)
In the first one, take a, b, c out of the three rows. In the second, move the column of ones to the front by two column interchanges (each changes the sign, so the sign is unchanged overall). Writing
\( \displaystyle V = \begin{vmatrix}1 & a & a^{2}\\ 1 & b & b^{2}\\ 1 & c & c^{2}\end{vmatrix} \)
we get
\(\Delta = abc\,V - V = (abc - 1)V\)
Now \(V = (b - a)(c - a)(c - b)\), which is not zero because a, b, c are distinct. So \(\Delta = 0\) gives
\(abc - 1 = 0\)
\(abc = 1\)
Check with \(a = 1,\ b = 2,\ c = \frac{1}{2}\) (so \(abc = 1\)): the rows are \((1, 1, 0)\), \((2, 4, 7)\), \(\left(\frac{1}{2}, \frac{1}{4}, -\frac{7}{8}\right)\), and the determinant is \(1\left(-\frac{7}{2} - \frac{7}{4}\right) - 1\left(-\frac{7}{4} - \frac{7}{2}\right) + 0 = 0\).
Why the other options are incorrect
Option (1) \(abc = -1\) — Incorrect. Then \(\Delta = (-1 - 1)V = -2V \ne 0\).
Option (3) \(abc = 0\) — Incorrect. Then \(\Delta = (0 - 1)V = -V \ne 0\).
Option (4) \(a + b + c = 0\) — Incorrect. For example a = 1, b = 2, c = −3 gives \(abc = -6\), so \(\Delta = -7V \ne 0\).