<s>[INST] Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$ Factorise the determinant $\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}$.
My textbook only provides two simple examples.
Really have no idea how to do this type of questions..
 [/INST] Regard $b$ and $c$ as constants, and the determinant as a polynomial in $a$. Then find ways of making the determinant equal to 0, and by the Factor Theorem you'll get a factor of the determinant.
Obvious choices: set $a=b$ or $a=c$ and you'll have two identical rows, so $(a-b)$ and $(a-c)$ are factors.
We can clearly see that permuting the variables is the same as permuting the rows, and hence only changes the determinant by a sign. Hence permuting the variables in any factor gives another factor, so $(b-c)$ is a factor.
What's left? Well, we can see without much difficulty that the determinant is cubic in $a$, and the coefficient of $a^3$ is $(b-c)$. We've already taken out a factor of $(b-c)$ and two monic linear factors, so whatever's left is linear and monic in $a$ and, by similar arguments, must also be so in $b$ and $c$. Hence it pretty much has to be $a + b + c + k$ for some constant $k$. So we've got as far as the following:
$$\det\begin{pmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{pmatrix}=(a-b)(a-c)(b-c)(a+b+c+k)$$
for some $k$, and all we need to do is find $k$. I first tried setting $b=c=0$ but that just leaves you with $0k = 0$, so instead let's set $b=1$, $c=0$:
$$\begin{align}
\det\begin{pmatrix} a^3+a^2 & a & 1 \\ 2 & 1 & 1 \\ 0 & 0 &1\end{pmatrix}&=a(a-1)(a+1+k) \\
\det\begin{pmatrix} a^3+a^2 & a \\ 2 & 1\end{pmatrix}&=a(a-1)(a+1+k) \\
a^3 + a^2 - 2a &=a(a-1)(a+1+k) \\
a^2 + a - 2 &= (a-1)(a+1+k)\end{align}$$
Comparing constant terms, $-2 = -1-k$ so $k=1$, and we're done!
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