Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. 

Then finding the absolute value of the matrix newly formed.

Also, 

C_ij = (–1)^i+j × M_ij

A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\) 

⇒ M₁₁ = \(\begin{bmatrix} b&ca \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M₁₁ = b × ab – c × ca 

M₁₁ = ab² – ac²

⇒ M₂₁ = \(\begin{bmatrix} a&bc \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M₂₁ = a × ab – c × bc 

M₂₁ = a²b – c²b

⇒ M₃₁ = \(\begin{bmatrix} a&bc \\[0.3em] b & ca \\[0.3em] \end{bmatrix}\)

M₃₁ = a × ca – b × bc 

M₃₁ = a²c – b²c 

C₁₁ = (–1)¹⁺¹ × M₁₁ 

= 1 × (ab² – ac²) 

= ab² – ac² 

C₂₁ = (–1)²⁺¹ × M₂₁ 

= –1 × (a²b – c²b) 

= c²b – a²b 

C₃₁ = (–1)³⁺¹ × M₃₁ 

= 1 × (a²c – b²c) 

= a²c – b²c 

Now expanding along the first column we get 

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ 

= 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c) 

= ab² – ac² + c²b – a²b + a²c – b²c