|

1	x	x2
x2	1	x
x	x2	1

|
Applying R1 → R1+R2+R3, we have:
Δ =
= 1 + x + x2 |

1	1	1
x2	1	x
x	x2	1

|
Applying C2 → C2−C1 and C3 → C3−C1, we have:
Δ = 1 + x + x2 |

1	0	0
x2	1−x2	x−x2
x	x2−x	1−x

|
= (1 + x + x2) (1 - x) (1 - x) |

1	0	0
x2	1+x	x
x	−x	1

|
= (1 - x3) (1 - x) |

1	0	0
x2	1+x	x
x	−x	1

|
Expanding along R1, we have:
Δ = (1 - x3) (1 - x) (1) |

1+x	x
−x	1

|
= (1 - x3) (1 - x) (1 + x + x2)
= (1 - x3) (1 - x2)
= (1−x3)2
Hence proved.