Similar Vandermonde

The proof is similar to the proof of the Vandermonde determinant. Let $P(x) = (x-a)(x-b)(x-c)(x+a+b+c).$ Then $P(x) = x^4 + a_2x^2 + a_1x + a_0$ for some $a_i.$ Then, since adding appropriate multiples of the first three rows to the fourth row doesn't change the determinant, we get $\left| \begin{matrix} 1&1&1&1 \\ a&b&c&d \\ a^2&b^2&c^2&d^2 \\ a^4&b^4&c^4&d^4 \end{matrix} \right| = \left| \begin{matrix} 1&1&1&1 \\ a&b&c&d \\ a^2&b^2&c^2&d^2 \\ P(a)&P(b)&P(c)&P(d) \end{matrix} \right| = \left| \begin{matrix} 1&1&1&1 \\ a&b&c&d \\ a^2&b^2&c^2&d^2 \\ 0&0&0&P(d) \end{matrix} \right|.$ So the determinant is $P(d)$ times the determinant of the $3\times 3$ upper left-hand corner, which is a Vandermonde determinant, equal to $(c-b)(c-a)(b-a),$ and $P(d) = (d-a)(d-b)(d-c)(a+b+c+d),$ so putting it together gives $(c-b)(c-a)(b-a)(d-a)(d-b)(d-c)(a+b+c+d) = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d).$

By subtracting the first column from the second, we see that $a-b$ is a factor of the determinant $\Delta$ . By considering all pairs of columns, we deduce that $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$ is a factor of $\Delta$ .

It is clear that $\Delta$ is a homogeneous polynomial of degree $7$ in $a,b,c,d$ , and hence $\Delta = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)L$ where $L$ is a homogeneous linear polynomial in $a,b,c,d$ . By comparing the coefficients of $bc^2d^4$ , $ac^2d^4$ , $ab^2d^4$ and $ab^2c^4$ in the determinant, it is easy to show that $L = a+b+c+d$ , and hence $\Delta = (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)(a+b+c+d)$

This determinant has a very nice symmetry of variables in a,b,c,d. Putting any 2 equal, tge determinant turns 0. Thus Π(a-b) divides the determinant where Π(a-b) is product of all distinct differences of pair of variables. Notice that the product has a degree of 6 whereas the determinant has a degree of 7. Thus we only missed out a linear factor which is symmetric in a,b,c,d. The only possibility is a+b+c+d. Thus our determinant is (a+b+c+d)Π(a-b). Now just put a=1,b=0,c=-1,d=2 and check whether both expression match or not. In case they don't just negate any one of the difference in the product of determinant that we derived.

Using the row reduction method, we can arrive at the result simply.