When presented with the following:
Use row or column operations to find the determinant in factored form:
$\left\vert \begin{array}{llll} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \\ \end{array} \right|$
I don't understand how I could pull a factor from this in a meaningful way using only row and column operations.
Then I'll just get the determinant in terms of $\dfrac{1}{a} ,\ \dfrac{b}{a}, ...$ etc.
What am I missing?
Perhaps the problem is just poorly worded, because it should specify in what way to determine this value (either exclusively or non-exclusively). But, if exclusively, I am at a loss.
Let's have a look at this inductively: start with the easy $$ d_2=\begin{vmatrix} 1 & 1 \\ a & b \end{vmatrix} = (b-a). $$ Subtracting the first column from the second, we can write this as $$ d_2=\begin{vmatrix} 1 & 1 \\ a & b \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ a & b-a \end{vmatrix}. $$ Now suppose we have $n$ columns: $$ d_n=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_n & a_{n-1} & \cdots & a_1 \\ a_n^2 & a_{n-1}^2 &\cdots & a_1^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_n^{n-1} & a_{n-1}^{n-1} & \cdots & a_1^{n-1} \end{vmatrix}. $$ One thing is easy to see: if $a_j=a_k$ for some $j \neq k$, the determinant is zero. Hence it has a factor $(a_j-a_k)$. It will turn out that's all of them (you can see this by counting powers of $a_k$, but that doesn't give you the constant). Subtract the first column from the others: $$ d_n=\begin{vmatrix} 1 & 0 & \cdots & 0 \\ a_n & a_{n-1}-a_n & \cdots & a_1-a_n \\ a_n^2 & a_{n-1}^2-a_n^2 &\cdots & a_1^2-a_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_n^{n-1} & a_{n-1}^{n-1}-a_n^{n-1} & \cdots & a_1^{n-1}-a_n^{n-1} \end{vmatrix} $$ Now we can expand by the first row: $$ d_n=\begin{vmatrix} a_{n-1}-a_n & \cdots & a_1-a_n \\ a_{n-1}^2-a_n^2 &\cdots & a_1^2-a_n^2 \\ \vdots & \ddots & \vdots \\ a_{n-1}^{n-1}-a_n^{n-1} & \cdots & a_1^{n-1}-a_n^{n-1} \end{vmatrix} $$ This can be factorised, using $a^k-b^k=(a-b)(a^{k-1}+a^{k-2}b+\dotsb+a^{k-r-1}b^r +b^{k-1})$, and then every column has a factor $a_k-a_n$, so $$ d_n=\begin{vmatrix} 1 & \cdots & 1 \\ a_{n-1}+a_n &\cdots & a_1+a_n \\ \vdots & \ddots & \vdots \\ a_{n-1}^{n-2}+a_{n-1}^{n-3}a_n+\dotsb+a_n^{n-2} & \cdots & a_1^{n-2}+a_1^{n-3}a_n+\dotsb+a_n^{n-2} \end{vmatrix} \prod_{i=1}^{n-1} (a_i-a_n) $$ Now, we could just subtract the first column again, but that looks horrendous! The trick is to notice that the $k$th row is of the form $$ \mathbf{a}^{k-1} + a_n \mathbf{a}^{k-2} + \dotsb + a_n^{k-1} \mathbf{a}^{0},$$ where $\mathbf{a}^{k}=(a_{n-1}^k,\dotsc,a_{1}^k)$. Hence we can subtract $a_n$ times the first row from the second row, and $a_n^2$ times the first row from the third row, and so on, leaving $\mathbf{a}$ as the second row: $$ d_n=\begin{vmatrix} 1 & \cdots & 1 \\ a_{n-1} &\cdots & a_1 \\ \vdots & \ddots & \vdots \\ a_{n-1}^{n-2}+a_{n-1}^{n-3}a_n+\dotsb+a_{n-1} a_{n}^{n-3} & \cdots & a_1^{n-2}+a_1^{n-3}a_n+\dotsb+a_1 a_n^{n-3} \end{vmatrix} \prod_{i=1}^{n-1} (a_i-a_n) $$ and then $a_n$ times the second row from the third row, and so on and so on (basically, we're getting rid of the multiples of $\mathbf{a}^k$ in each row but the first in which it appears). And so we end up with $$ d_n=\begin{vmatrix} 1 & 1 & \cdots & 1 \\ a_{n-1} & a_{n-2} & \cdots & a_1 \\ a_{n-1}^2 & a_{n-2}^2 &\cdots & a_1^2 \\ \vdots & \vdots & \ddots & \vdots \\ a_{n-1}^{n-2} & a_{n-2}^{n-2} & \cdots & a_1^{n-2} \end{vmatrix} \prod_{i=1}^{n-1}(a_i-a_n), $$ and then applying induction, $$ d_n = \prod_{i<j} (a_i-a_j). $$