Question

If a, b, c are all different and if $\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ then $-abc=$

Answer 1

Let $D=\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|$
$D=\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|$
$=\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
$=(-1{)}^1\left|\begin{array}{ccc}1& a^2& a\\ 1& b^2& b\\ 1& c^2& c\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|[C_1\leftrightarrow C_3$ in 1st det.]
$=(-1{)}^2\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|[C_2\leftrightarrow C_3$ in 1st det.]
$=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
$=(1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$
$=(1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}\right|[R_2\to R_2-R_1$ and $R_3\to R_3-R_1]$
$=(1+abc)\left|\begin{array}{cc}b-a& b^2-a^2\\ c-a& c^2-a^2\end{array}\right|$ (expanding along 1st row)
$=(1+abc)(b-a)(c-a)\left|\begin{array}{cc}1& b+a\\ 1& c+a\end{array}\right|$
$=(1+abc)(b-a)(c-a)(c+a-b-a)$
$=(1+abc)(b-a)(c-a)(c-b)$
$\therefore D=(1+abc)(a-b)(b-c)(c-a)$
But given $D=0$
$\therefore (1+abc)(a-b)(b-c)(c-a)=0$
$\therefore 1+abc=0$ [since a, b, c are different $\because a\ne b,b\ne c,c\ne a$]
Hence $-abc=1$