the value of a determinant

Question

Let be a polynomial function $P\in \mathbb{R}[X]$.If I divide $P$ by $(x-1)(x-2)(x-3)(x-4)$ I get a remainder without "free term" ( like $ax^{3}+bx^{2}+cx$ )

I have to calculate the determinant:

$$\begin{vmatrix} P(1) & 1 & 1 &1 \\ P(2) & 2 & 4 &8 \\ P(3) & 3 & 9 &27 \\ P(4) & 4 & 16 &64 \end{vmatrix}$$

My try: I wrote that $P(x)=(x-1)(x-2)(x-3)(x-4)\cdot Q(x)+ax^{3}+bx^{2}+cx$

So for $x=1=> P(1)=a+b+c$

$x=2=> P(2)=8a+4b+2c$

$x=3=> P(3)=27a+9b+3c$

$x=4=> P(4)=64a+16b+4c$

And now I just have to replace the results in my determinant but it takes me a lot of time to solve the determinant.I'm wondering if there is a short way to solve this. Can you help me with some ideas?

Answer 1

Hint: Take column one and subtract $c$ from column $2,$ $b$ from column 3 and $a$ of column 4. Do you get a column full of $0's$? What is the determinant of that?

Answer 2

Hint: What is $a$ times the fourth column, plus $b$ times the third column, plus $c$ times the second?