Δ = |

1	1	1
a	b	c
a3	b3	c3

|
Applying C1 → C1−C3 and C2 → C2−C3, we have:
Δ = |

1−1	1−1	1
a−c	b−c	c
a3−c3	b3−c3	c3

|
=
= (c - a) (b - c)
Applying C1 → C1+C2, we have:
Δ = (c - a) (b - c)
= (b - c) (c - a) (a - b)
= (a - b) (b - c) (c - a) (a + b + c)
Expanding along C1, we have:
Δ = (a - b) (b - c) (c - a) (a + b + c) - 1 |

0	1
1	c

|
= (a - b) (b - c) (c - a) (a + b + c)
Hence proved.