1.

Prove that `|{:(1, 1, 1),(a, b, c),(a^(3), b^(3), c^(3)):}|=(a-b)(b-c)(c-a)(a+b+c)`

Answer» Let the value of the given determinant be `Delta`. Then,
`Delta =|{:(1, 1, 1),(a, b, c),(a^(3), b^(3), c^(3)):}|`
`=|{:(0, 0, 1),(a-c, b-c, c),(a^(3)-c^(3), b^(3)-c^(3), c^(3)):}|" "["applying"C_(1) to (C_(1) -C_(3))" and " C_(2) to (C_(2) -C_(3))]`
`=(a-c)(b-c) *|{:(0, 0, 1),(1, 1, c),(a^(2)+ac+c^(2), b^(2)+bc+c^(2), c^(3)):}|" "["taking out (a-c) and (b-c) common from"C_(1) " and " C_(2)]` ltbr `=(a-c)(b-c) *1*|{:(1, 1),(a^(2)+ac+c^(2), b^(2) + ac +c^(2)):}| " "["expanded by"R_(1)]`
`=(a-c)(b-c) * [(b^(2) +bc+c^(2)) -(a^(2) +ac+c^(2))]`
`=(a-c)(b-c)[(b^(2)-a^(2)) + (b-a)c]`
`= (a-c)(b-c)(b-a)(b+a+c)`
`=(a-b)(b-c)(c-a)(a+b+c)`
Hence, `Delta = (a-b)(b-c)(c-a) (a+b+c)`.


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