$$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any success.
Best Answer
If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get $$ \begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}. $$ Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one, and 3 times the 2$^{\rm nd}$ column from the 4$^{\rm th}$ one: $$ \begin{vmatrix}a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6\end{vmatrix}=0. $$ Now it is clear that the determinant is zero, as the fourth column is a multiple of the third one (or, factor 3 from the fourth column to get two equal columns).