By applying R₁→ R₁ + R₂ + R₃, we get

Taking (a + b +c) common from the first row, we get

By applying C₂→ C₂ – C₁, we get

By applying C₃→ C₃ – C₁, we get

Now, expanding along first row, we get

= (a + b+ c)[1×{-(a + b + c)×{-(a + b + c)} – 0}]

= (a + b + c)[(a + b + c)²]

= (a + b + c)³