Polynomials are Awesome

$\omega$ represents cube root of unity. Why not we let, for the sake of solving the problem in a simple manner, that cube root be $1$ itself?

The problem thus becomes much simpler. We just need to find $p(1)$ , being given $p(2)$ , $p(3)$ , $p(4)$ , and $p(5)$ .

We can straightforwardly apply the Method Of Differences .

We make the difference table as follows:-

$\begin{array} {l l l l l l l} n & f(n) & D_1(n) & D_2(n) & D_3(n) & D_4(n) \\ 1 & 1 & -56 & -70 & 0 & 24 \\ 2 & -55 & -126 & -70 & 24 \\ 3 & -181 & -196 & -46 \\ 4 & -377 & -242 \\ 5 & -619\\ \end{array}$

We hereby use the fact that being the polynomial quartic, the $4^{th}$ difference will be constant, and will be equal to $4!=24$ times the leading coefficient, being $1$ .

Let $p(x) = x^4 + ax^3 + bx^2 + cx + d$

Substituting $2,3,4$ and $5$ into $x$ , we obtain four simultaneous equations.

Solving them yields $a=-10, \ b=0, \ c=-1$ and $d=11$

Hence,

$f(\omega)=\omega^4-10\omega^3-\omega+11$

$\qquad = \omega-10-\omega+11$

$\qquad = \large{1}$

Let $p(x) = x^{4}+bx^{3}+cx^{2}+dx+e$ .
$p(2) = 16+8b+4c+2d+e$ .
$p(3) = 81+27b+9c+3d+e$ .
$p(4) = 256+64b+16c+4d+e$ .
$p(5) = 625+125b+25c+5d+e$ .

By solving these equations, we will get $b = -10$ , $c = 0$ , $d = -1$ and $e = 11$ .
Hence, $p(x) = x^{4}-10x^{3}-x+11$ .
$p(\omega) = \omega^4-10\omega^3-\omega+11$ = $\omega-10+\omega+11 = 1$ .

Add (1, 1) to (2, -55), (3, -181), (4, -377) and (5, -619) to form a monic polynomial of degree 4, p(x) = x^4 - 10 x^3 - x+ 11, by Cramer's 5 x 5 determinants.

x^3 - 1 = (x - 1)(x^2 + x + 1) for omega, w = -1/ 2 + j Sqrt (3)/ 2,

w^4 = w <=> w^4 - w = 0, while w^3 = 1,

p (w) = w^4 - 10 w^3 - w + 11 = - 10 + 11 = 1

Note:

D = 288

D 1 = 288

D p = -2880

D q = 0

D r = -288

D s = 3168