TITLE: What will be the value of the following determinant without expanding it?
QUESTION [27 upvotes]: $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.

REPLY [6 votes]: If you expand the squares, you'll see that every column is a linear combination of
$$
\begin{bmatrix}a^2 \\ b^2 \\ c^2 \\ d^2\end{bmatrix},
\begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}, \text{ and }
\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}.
$$
There are four columns that are linear combinations of three vectors. So they can't be linearly independent, thus the determinant is $0$.
(Or: The column space of the matrix is spanned by $3$ vectors. So the rank of the matrix is at most $3$, which means the matrix is singular. So the determinant is $0$.)