Δ = |

a	b	c
a−b	b−c	c−a
b+c	c+a	a+b

|
Applying C1 → C1+C2+C3
Δ = |

a+b+c	b	c
0	b−c	c−a
2(a+b+c)	c+a	a+b

|
Δ = (a + b + c) |

1	b	c
0	b−c	c−a
0	c+a−2b	a+b−2c

|
R3 → R3−2R1
Δ = (a + b + c) |

1	b	c
0	b−c	c−a
0	c+a−2b	a+b−2c

|
Expanding along C1, we have,
Δ = (a +b +c) ((b –c) (a + b – 2c) – (c – a) (c + a – 2b))
⇒ Δ = (a + b + c) ((ba + b2 - 2bc - ca - cb + 2c2 - (c2 + ac - 2bc - ac - a2 + 2ab))
⇒ Δ = (a + b + c) (a2+b2+c2 - ca - bc - ab)
⇒ Δ = (a + b + c) (a2+b2+c2 - ab - bc - ac)
⇒ Δ = a3+b3+c3 - 3abc = R.H.S.