Explanation

(a)
Given determinant, $\mathrm{\Delta} \equiv \left| \begin{matrix}
a & a^{2} & a^{3} + 1 \\
b & b^{2} & b^{3} + 1 \\
c & c^{2} & c^{3} + 1
\end{matrix} \right| = 0$
On splitting the determinant into two determinants, we get
$\mathrm{\Delta} \equiv abc\left| \begin{matrix}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{matrix} \right| + \left| \begin{matrix}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{matrix} \right| = 0$
$\Rightarrow (1 + abc)\left\lbrack 1\left( bc^{2} - cb^{2} \right) - a\left( c^{2} - b^{2} \right) + a^{2}(c - b) \right\rbrack = 0$
$\Rightarrow (1 + abc)\left\lbrack (a - b)(b - c)(c - a) \right\rbrack = 0$
Since $a,b,c$ are different, the second factor cannot be zero.
Hence, $1 + abc = 0$