Consider the following determinant: [Tex]\Delta = \begin{vmatrix} 1 & a & bc…

Concept: A determinant is an alternating function of its rows — swapping two rows negates its value. So if setting two variables equal makes two rows of a determinant identical, the determinant must vanish identically at that substitution, which means the corresponding difference of those variables is a factor of the fully expanded determinant. This factor-theorem-for-determinants idea lets us predict linear factors of Δ before fully expanding it.

Application:

1. Expand Δ along the first row: Δ = 1·(b·ab − ca·c) − a·(1·ab − ca·1) + bc·(1·c − b·1).

2. Simplify each product: Δ = ab2 − ac2 − a2b + a2c + bc2 − b2c.

3. Group the six terms in pairs sharing a common factor: Δ = (b − c)(ab + ac − a2 − bc).

4. Factor the remaining quadratic: ab + ac − a2 − bc = −(a − b)(a − c).

5. Substitute back: Δ = (b − c)·[−(a − b)(a − c)] = (a − b)(b − c)(c − a).

6. So the fully factorized determinant has three linear factors: (a − b), (b − c), and (c − a) — matching the option a − b.

Cross-check: Substituting a = b directly in the original determinant makes the first two rows identical — both become (1, a, ac) — which forces Δ = 0, independently confirming that (a − b) is a genuine factor.