If \( \alpha,\beta,\gamma \) are the roots of \( x^3+mx+n=0 \) then the value of \( \begin{vmatrix}1+\alpha & 1 & 1\\1 & 1+\beta & 1\\1 & 1 & 1+\gamma\end{vmatrix} \) is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( m-n \)
For \( x^3+0\cdot x^2+mx+n=0 \), the relations between roots and coefficients give
\( \displaystyle \alpha+\beta+\gamma=0 \)
\( \displaystyle \alpha\beta+\beta\gamma+\gamma\alpha=m \)
\( \displaystyle \alpha\beta\gamma=-n \)
Expanding the determinant along the first row:
\( \displaystyle \Delta=(1+\alpha)\left[(1+\beta)(1+\gamma)-1\right]-1\left[(1+\gamma)-1\right]+1\left[1-(1+\beta)\right] \)
\( \displaystyle =(1+\alpha)(\beta+\gamma+\beta\gamma)-\gamma-\beta \)
\( \displaystyle =\beta+\gamma+\beta\gamma+\alpha\beta+\alpha\gamma+\alpha\beta\gamma-\gamma-\beta \)
\( \displaystyle =(\alpha\beta+\beta\gamma+\gamma\alpha)+\alpha\beta\gamma \)
So
\( \displaystyle \Delta=m+(-n)=m-n \)
Why the other options are incorrect
Option (1) \( m+n+2 \) — Incorrect. The determinant has no constant term: \( \Delta=(\alpha\beta+\beta\gamma+\gamma\alpha)+\alpha\beta\gamma=m-n \), so 2 cannot appear.
Option (2) \( m-n-2 \) — Incorrect. The determinant has no constant term, so −2 cannot appear.
Option (3) \( m+n \) — Incorrect. This takes \( \alpha\beta\gamma=n \); for a cubic \( x^3+\dots+n \), the product of the roots is −n.