If Delta_1=left|begin{array}{lll}1 & a^2 & a^ | vedclass

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(A) First,we evaluate $\Delta_1 = \left|\begin{array}{lll}1 & a^2 & a^3 \ 1 & b^2 & b^3 \ 1 & c^2 & c^3\end{array}ight|$.Applying $R_2 ightarrow

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$a b+b c+c a$

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(A) First,we evaluate $\Delta_1 = \left|\begin{array}{lll}1 & a^2 & a^3 \ 1 & b^2 & b^3 \ 1 & c^2 & c^3\end{array}ight|$.Applying $R_2 ightarrow R_2-R_1$ and $R_3 ightarrow R_3-R_1$:$\Delta_1 = \left|\begin{array}{ccc}1 & a^2 & a^3 \ 0 & b^2-a^2 & b^3-a^3 \ 0 & c^2-a^2 & c^3-a^3\end{array}ight| = (b-a)(c-a) \left|\begin{array}{ccc}1 & a^2 & a^3 \ 0 & b+a & b^2+a^2+ab \ 0 & c+a & c^2+a^2+ac\end{array}ight|$.Expanding along the first column:$\Delta_1 = (b-a)(c-a) [(b+a)(c^2+a^2+ac) - (c+a)(b^2+a^2+ab)]$$= (b-a)(c-a) [bc^2+a^2b+abc+ac^2+a^3+a^2c - (cb^2+ca^2+abc+ab^2+a^3+a^2b)]$$= (b-a)(c-a) [bc^2+ac^2-cb^2-ab^2+a^2c-a^2b]$$= (b-a)(c-a) [bc(c-b) + a^2(c-b) + a(c^2-b^2)]$$= (b-a)(c-a)(c-b) [bc + a^2 + a(c+b)]$$= -(a-b)(b-c)(c-a) (ab+bc+ca)$.Now,we evaluate $\Delta_2 = \left|\begin{array}{lll}bc & b+c & 1 \ ca & c+a & 1 \ ab & a+b & 1\end{array}ight|$.Applying $R_2 ightarrow R_2-R_1$ and $R_3 ightarrow R_3-R_1$:$\Delta_2 = \left|\begin{array}{ccc}bc & b+c & 1 \ ca-bc & a-b & 0 \ ab-bc & a-c & 0\end{array}ight| = \left|\begin{array}{ccc}bc & b+c & 1 \ -c(b-a) & -(b-a) & 0 \ -b(c-a) & -(c-a) & 0\end{array}ight|$.Taking $(b-a)$ from $R_2$ and $(c-a)$ from $R_3$:$\Delta_2 = (b-a)(c-a) \left|\begin{array}{ccc}bc & b+c & 1 \ -c & -1 & 0 \ -b & -1 & 0\end{array}ight| = (b-a)(c-a) [1(c-b)] = -(a-b)(b-c)(c-a)$.Finally,$\frac{\Delta_1}{\Delta_2} = \frac{-(a-b)(b-c)(c-a)(ab+bc+ca)}{-(a-b)(b-c)(c-a)} = ab+bc+ca$.

If $\Delta_1=\left|\begin{array}{lll}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{array}\right|$ and $\Delta_2=\left|\begin{array}{lll}b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1\end{array}\right|$,then $\frac{\Delta_1}{\Delta_2}=$

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