Prove the given :-

Apply C₁ -> C₁ - C₃ and C₂ -> C₂ - C₃

 

On simplifying the so obtained determinant, you will get,

              |(b+c-a)         0              a²|

(a+b+c)² |0             (c+a-b)           b²|

              |(c-a-b)     (c-a-b)     (a+b)²|

Apply R₃ -> R₃ - (R₁ + R₂)

And then apply C₁ -> aC₁+ C₃, C₂ -> bC₂ + C₃

On simplification, you will get the required result.