(i)$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right| = (a - b)(b - c)(c - a).$
(ii) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right| = (a - b)(b - c)(c - a)(a + b + c)$

(i) L.H.S. $=$ $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&a&{{a^2}}\\1&b&{{b^2}}\\1&c&{{c^2}}\end{array}} \right|$

Applying ${R_1} \to {R_1} - {R_2}$ and ${R_2} \to {R_2} - {R_3},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{a - b}&{{a^2} - {b^2}}\\0&{b - c}&{{b^2} - {c^2}}\\1&c&{{c^2}}\end{array}} \right|$

$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{(a - b)}&{(a - b)(a + b)}\\0&{(b - c)}&{(b - c)(b + c)}\\1&c&{{c^2}}\end{array}} \right|$

Taking out $(a - b)$ and $(b - c)$ common from${R_1}$ and${R_2}$ respectively,

we get
$(a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&1&{a + b}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right|$

Applying ${R_1} \to {R_1} - {R_2},$

we get
$(a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&{a - c}\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right|$

Taking out $(a - c)$ common from ${R_1},$

we get
$(a - b)(b - c)(a - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\0&1&{b + c}\\1&c&{{c^2}}\end{array}} \right|$

Expanding along ${R_1},$

we get
$(a - b)(b - c)(a - c)1(0 - 1) = (a - b)(b - c)(c - a)$

$=$ R.H.S.
Hence, proved.

(ii) L.H.S. $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}} \right|$

Applying ${C_1} \to {C_1} - {C_2}$ and ${C_2} \to {C_2} - {C_3},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\{a - b}&{b - c}&c\\{{a^3} - {b^3}}&{{b^3} - {c^3}}&{{c^3}}\end{array}} \right|$

Taking out $(a - b)$ and $(b - c)$ common from ${C_1}$

and ${C_2}$ respectively,

we get
$(a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\1&1&c\\{{a^2} + ab + {b^2}}&{{b^2} + {c^2} + bc}&{{c^3}}\end{array}} \right|$

Applying ${C_1} \to {C_1} - {C_2},$

we get
$(a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\0&1&c\\{{a^2} + ab + {b^2} - {b^2} - {c^2} - bc}&{{b^2} + {c^2} + bc}&{{c^3}}\end{array}} \right|$

$= (a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\0&1&c\\{{a^2} - {c^2} + b(a - c)}&{{b^2} + {c^2} + bc}&{{c^3}}\end{array}} \right|$

$= (a - b)(b - c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\0&1&c\\{(a - c)(a + b + c)}&{{b^2} + {c^2} + bc}&{{c^3}}\end{array}} \right|$

Taking out $(a - c) \times (a + b + c)$ common from ${C_1}$,

we get
$(a - b)(b - c)(a - c)(a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&0&1\\0&1&c\\1&{{b^2} + {c^2} + bc}&{{c^3}}\end{array}} \right|$

Expanding along ${R_1}$,

we get
$(a - b)(b - c)(a - c)(a + b + c)( - 1)$

$= (a - b)(b - c)(c - a)(a + b + c) =$ R.H.S.
Hence, proved.