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.

$$\text{Det}(A)=\begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix}=\begin{vmatrix}1& a & a^2 & a^4 \\ 0 & b-a & b^2-a^2 & b^4-a^4 \\ 0 & c-a & c^2-a^2 & c^4-a^4 \\0 & d-a & d^2-a^2 &d^4-a^4 \end{vmatrix}=\begin{vmatrix}   b-a & b^2-a^2 & b^4-a^4 \\ c-a & c^2-a^2 & c^4-a^4 \\ d-a & d^2-a^2 &d^4-a^4 \end{vmatrix}$$
thus
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix}   1 & b+a &(b+a)(b^2+a^2) \\ 1 & c+a & (c+a)(c^2+a^2) \\ 1 & d+a &(d+a)(d^2+a^2) \end{vmatrix}$$
As a result
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix}  1 & b &(b+a)(b^2+a^2) \\ 0 & c-b & (c+a)(c^2+a^2)-(b+a)(b^2+a^2) \\ 0 & d-b &(d+a)(d^2+a^2)-(b+a)(b^2+a^2) \end{vmatrix}$$
therefore
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix}   c-b & (c+a)(c^2+a^2)-(b+a)(b^2+a^2) \\  d-b &(d+a)(d^2+a^2)-(b+a)(b^2+a^2) \end{vmatrix}$$
In other words
$$\text{Det}(A)=(b-a)(c-a)(d-a)\begin{vmatrix}   c-b & (c^3-b^3)+a(c^2-b^2)+a^2(c-b) \\  d-b &(d^3-b^3)+a(d^2-b^2)+a^2(d-b) \end{vmatrix}$$
or
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)\begin{vmatrix}   1 & (c^2+bc+b^2)+a(c+b)+a^2 \\  1 &(d^2+bd+b^2)+a(d+b)+a^2 \end{vmatrix}$$
Finally we have
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)\begin{vmatrix}   1 & (c^2+bc+b^2)+a(c+b)+a^2 \\  0 &(d^2-c^2)+b(d-c)+a(d-c) \end{vmatrix}\\
\quad=(b-a)(c-a)(d-a)(c-b)(d-b)[(d^2-c^2)+b(d-c)+a(d-c)]\\
$$
or
$$\text{Det}(A)=(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)(a+b+c+d)$$
