<s>[INST] If det $A = 0$ and $\det B \neq 0$ then show that $abc = -1$ This has been hurting my head for a while now....
If
$$
\det\begin{bmatrix}a&a^2&1+a^3\\b&b^2&1+b^3\\c&c^2&1+c^3\end{bmatrix}=0
$$
And
$$
\det\begin{bmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{bmatrix} ≠0
$$
Then show that $abc=-1$.
 [/INST] First note that,
$$\begin{vmatrix}
a & a^2 & a^3+1 \\
b & b^2 & b^3+1 \\
c & c^2 & c^3+1 \\
\end{vmatrix}
= 
\begin{vmatrix}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3 \\
\end{vmatrix}
+
\begin{vmatrix}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1 \\
\end{vmatrix}=0$$
Then, $$abc\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}
=
\begin{vmatrix}
a & a^2 & a^3 \\
b & b^2 & b^3 \\
c & c^2 & c^3 \\
\end{vmatrix}
=
-\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2 \\
\end{vmatrix}$$
Therefore, $abc=-1$
</s>