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Problem 30: Show that

\begin{align*} \Delta =\left| \begin{array}{ccc} 1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\\ \end{array}\right| = (a-b)(b-c)(c-a). \end{align*}

Problem 28: Show that

\begin{align*} \Delta =\left| \begin{array}{ccc} 1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\\ \end{array}\right| = (a-b)(b-c)(c-a). \end{align*}

Solution:

Applying$R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$

\begin{align*} \Delta =\left| \begin{array}{ccc} 1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\\ \end{array}\right| \end{align*} on expanding along the $C_1$ \begin{align*} \Delta =1\left| \begin{array}{ccc} b-a& b^2-a^2\\ c-a& c^2-a^2\\ \end{array}\right|\\ \Delta =1\left| \begin{array}{ccc} b-a& (b-a)(b+a)\\ c-a& (c-a)(c+a)\\ \end{array}\right| \end{align*}

taking $(b-a)$ common from $R_1$ and $c-a$ common from $R_2$

\begin{align*} \Delta =(b-a)(c-a)\left| \begin{array}{ccc} 1& (b+a)\\ 1& (c+a)\\ \end{array}\right| \end{align*} \begin{align*} =\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\left\{\left(\mathrm{c}+\mathrm{a}\right)-(\mathrm{b}+\mathrm{a})\right\}\\ =\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\left(c-b\right)\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(c-a\right) \end{align*}

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