<s>[INST] How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint 

Replace $d$ by a variable $x$; make use of the fact that the sum of the roots of a fourth-degree polynomial is equal to the coefficient of $x^3$

but I didn't get that.
 [/INST] Consider instead the polynomial in $x$
$P(x)=\det\begin{pmatrix}1&x&x^2&x^3&x^4\\1&a&a^2&a^3&a^4\\1&b&b^2&b^3&b^4\\1&c&c^2&c^3&c^4\\1&d&d^2&d^3&d^4\end{pmatrix}$.
If you use Laplace's expansion in the first row, you'll notice that $P(x)$ has degree $4$. 
Also, $P(a)=P(b)=P(c)=P(d)=0$ because plugging $x=a,b,c,d$ creates two equal rows and the determinant vanishes.
Now, by Vieta's relations, and recalling that the terms in the Laplace's expansion alternate signs, the term in $x^3$ is the sum of the roots, so
$a+b+c+d=\frac{\det\begin{pmatrix}1& a& a^2& a^4\\1& b& b^2& b^4\\1& c& c^2& c^4\\1& d& d^2& d^4\end{pmatrix}}{\det\begin{pmatrix}1& a& a^2& a^3\\1& b& b^2& b^3\\1& c& c^2& c^3\\1& d& d^2& d^3\end{pmatrix}}$
Now we can use Vandermonde's determinant and finish the problem:
$\det\begin{pmatrix}1& a& a^2& a^4\\1& b& b^2& b^4\\1& c& c^2& c^4\\1& d& d^2& d^4\end{pmatrix}=(a+b+c+d)(d-a)(d-c)(d-b)(c-a)(c-b)(b-a)$.
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