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

*

*Take any sequence of consecutive square numbers. What do you know (or, if you don't know, try it: what do you notice) about the second differences?

*Each row of the matrix has four consecutive square numbers.* Note that column operations preserve the determinant in the same way that row operations do (this is obvious if we recall that the determinant of a matrix equals the determinant of its transpose, and column operations on your matrix are row operations on its transpose). One can fill the final three columns of your matrix with the first differences of your four consecutive square numbers, by applying the following column operations: subtract column 3 from column 4; subtract column 2 from column 3; subtract column 1 from column 2. (Why did we have to do it in this order?)

*Can you perform another set of column operations to difference these three first differences? The final two columns of your matrix will then be filled with the second differences of your original sequences of four consecutive square numbers. Recall the result from (1). What does this tell you about these two columns and hence the determinant?
$(*)$ Perhaps $a^2, \dots, (a+3)^2$ may not be "square numbers" in the sense that $a$ may not be a natural number. But this doesn't matter very much; the reason the second differences of the sequence $n^2, \, n\in\mathbb{N}$ are so nice is because of algebra that works just as well on $x^2, \, x\in \mathbb{R}$. In particular, what is $\left((x+2)^2-(x+1)^2\right) - \left((x+1)^2-x^2\right)$?

If you brush up a little on finite differences of higher polynomials you might want to have a think about how you could determine the following determinant, where each row has six consecutive fourth powers:
$$\begin{vmatrix}a^4 & (a+1)^4 & (a+2)^4 & (a+3)^4 & (a+4)^4 & (a+5)^4 \\ (b+6)^4 & (b+7)^4 & (b+8)^4 & (b+9)^4 & (b+10)^4 & (b+11)^4 \\ (c-3)^4 & (c-2)^4 & (c-1)^4 & c^4 & (c+1)^4 & (c+2)^4 \\ (d+20)^4 & (d+21)^4 & (d+22)^4 & (d+23)^4 & (d+24)^4 & (d+25)^4 \\ (e-8)^4 & (e-7)^4 & (e-6)^4 & (e-5)^4 & (e-4)^4 & (e-3)^4 \\ (f + 2016)^4 & (f+2017)^4 & (f+2018)^4 & (f+2019)^4 & (f+2020)^4 & (f+2021)^4 \end{vmatrix} $$
Expanding this out in full may not necessarily develop the situation to your advantage. But this time, with a fourth power polynomial, it isn't the second differences that come out the same...
