By using properties of determinants, show that:

Solution:

(i) Let $\Delta=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$.

Applying $R_{1} \rightarrow R_{1}-R_{3}$ and $R_{2} \rightarrow R_{2}-R_{3}$, we have:

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

Applying $R_{1} \rightarrow R_{1}+R_{2}$, we have:

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

Expanding along C1, we have:

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

Hence, the given result is proved.

(ii) Let $\Delta=\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$.

Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $C_{2} \rightarrow C_{2}-C_{3}$, we have:

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

Applying $C_{1} \rightarrow C_{1}+C_{2}$, we have:

$\begin{aligned} \Delta &=(c-a)(b-c)\left|\begin{array}{llll}0 & 0 & 1 \\ 0 & 1 & c \\ \left(b^{2}-a^{2}\right)+(b c-a c) & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array}\right| \\ &=(b-c)(c-a)(a-b) \mid \begin{array}{llll}0 & 0 & 1 \\ 0 & 1 & c \\ -(a+b+c) & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array} \\ &=(a-b)(b-c)(c-a)(a+b+c)\left|\begin{array}{lll}0 & 0 & 1 \\ -1 & \left(b^{2}+b c+c^{2}\right) & c^{3}\end{array}\right| \end{aligned}$

Expanding along C1, we have:

$\begin{aligned} \Delta &=(a-b)(b-c)(c-a)(a+b+c)(-1)\left|\begin{array}{ll}0 & 1 \\ 1 & c\end{array}\right| \\ &=(a-b)(b-c)(c-a)(a+b+c) \end{aligned}$

Hence, the given result is proved.