Discussion

Explanation:

Given, $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3-1\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3-1\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3-1\end{array}\right|$ = 0

⇒ $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3\end{array}\right|$ + $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& -1\\ \mathrm{b}& {\mathrm{b}}^2& -1\\ \mathrm{c}& {\mathrm{c}}^2& -1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& 1\\ \mathrm{b}& {\mathrm{b}}^2& 1\\ \mathrm{c}& {\mathrm{c}}^2& 1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ + $\left|\begin{array}{ccc}1& {\mathrm{a}}^2& \mathrm{a}\\ 1& {\mathrm{b}}^2& \mathrm{b}\\ 1& {\mathrm{c}}^2& \mathrm{c}\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

⇒ (abc - 1) × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

$\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ ≠ 0

∴ abc - 1 = 0

⇒ abc = 1

Hence, 1.

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