1.

`|[a,a^2,a^3-1],[b,b^2,b^3-1],[c,c^2,c^3-1]|=0 ` prove that `abc= I `

Answer» `|[a,a^2,a^3-1],[b,b^2,b^3-1],[c,c^2,c^3-1]| = 0`
`=>|[a,a^2,a^3],[b,b^2,b^3],[c,c^2,c^3]|-|[a,a^2,1],[b,b^2,1],[c,c^2,1]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-|[a,a^2,1],[b,b^2,1],[c,c^2,1]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-(-1)^2|[1,a,a^2],[1,b,b^2],[1,c,c^2]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-|[1,a,a^2],[1,b,b^2],[1,c,c^2]| = 0`
`=>|[1,a,a^2],[1,b,b^2],[1,c,c^2]|[abc-I] = 0`
As, `|[1,a,a^2],[1,b,b^2],[1,c,c^2]|` can not be `0`.
`:. abc - I = 0`
`=>abc = I.`


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