Let

We need to find the roots of Δ = 0.

Recall that the value of a determinant remains same if we apply the operation R_i→ R_i + kR_j or C_i→ C_i + kC_j.

Applying R₂→ R₂ – R₁, we get

Applying R₃→ R₃ – R₁, we get

Taking (b – x) and (c – x) common from R₂ and R₃, we get

Expanding the determinant along C₁, we have

Δ = (b – x)(c – x)(1)[(1)(c² + cx + x²) – (1)(b² + bx + x²)]

⇒ Δ = (b – x)(c – x)[c² + cx + x² – b² – bx – x²]

⇒ Δ = (b – x)(c – x)[c² – b² + cx – bx]

⇒ Δ = (b – x)(c – x)[(c – b)(c + b) + (c – b)x]

∴ Δ = (b – x)(c – x)(c – b)(c + b + x)

The given equation is Δ = 0.

⇒ (b – x)(c – x)(c – b)(c + b + x) = 0

However, b ≠ c according to the given condition.

⇒ (b – x)(c – x)(c + b + x) = 0

Case – I:

b – x = 0 ⇒ x = b

Case – II:

c – x = 0 ⇒ x = c

Case – III:

c + b + x = 0 ⇒ x = –(b + c)

Thus, b, c and –(b + c) are the roots of the given determinant equation.