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Sum of p(x) over the Roots of x^4 - x^3 - x^2 - 1 = 0 | JEE Advanced

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Question
Let α,β,γ,δ\alpha, \beta, \gamma, \delta be the roots of x4−x3−x2−1=0x^{4} - x^{3} - x^{2} - 1 = 0. Also consider
p(x)=x6−x5−x3−x2−x.p(x) = x^{6} - x^{5} - x^{3} - x^{2} - x .
 Then the value of p(α)+p(β)+p(γ)+p(δ)p(\alpha) + p(\beta) + p(\gamma) + p(\delta) cannot be
A44correct
B55correct
C66
D−1-1correct
Solution
Step 1: Use what a root satisfies. If xx is a root of x4−x3−x2−1=0x^{4} - x^{3} - x^{2} - 1 = 0, then
x4−x3−x2=1....(i)x^{4} - x^{3} - x^{2} = 1 . \qquad \text{...(i)}
 So the plan is to rewrite p(x)p(x) in terms of the block x4−x3−x2x^{4} - x^{3} - x^{2}. Step 2: Pull that block out of p(x)p(x). Multiplying (i)'s left side by x2x^{2} gives x6−x5−x4x^{6} - x^{5} - x^{4}, which supplies the two highest terms of p(x)p(x):
x2(x4−x3−x2)=x6−x5−x4.x^{2}\left(x^{4} - x^{3} - x^{2}\right) = x^{6} - x^{5} - x^{4} .
 Step 3: See what is still missing. Subtracting this from p(x)p(x),
p(x)−(x6−x5−x4)=x4−x3−x2−x,p(x) - \left(x^{6} - x^{5} - x^{4}\right) = x^{4} - x^{3} - x^{2} - x ,
 and the first three terms here are the block again. Hence
p(x)=(x2+1)(x4−x3−x2)−x.p(x) = \left(x^{2} + 1\right)\left(x^{4} - x^{3} - x^{2}\right) - x .
 Step 4: Evaluate at a root using (i).
p(x)=(x2+1)(1)−x=x2−x+1.p(x) = \left(x^{2}+1\right)(1) - x = x^{2} - x + 1 .
 Step 5: Add over all four roots.
∑p=∑α2−∑α+4.\sum p = \sum \alpha^{2} - \sum \alpha + 4 .
 Step 6: Get the symmetric sums from the quartic x4−x3−x2+0⋅x−1=0x^{4} - x^{3} - x^{2} + 0\cdot x - 1 = 0:
∑α=1,∑αβ=−1.\sum \alpha = 1, \qquad \sum \alpha\beta = -1 .
 Step 7: Use the identity ∑α2=(∑α)2−2∑αβ\sum\alpha^{2} = \left(\sum\alpha\right)^{2} - 2\sum\alpha\beta:
∑α2=12−2(−1)=3.\sum \alpha^{2} = 1^{2} - 2(-1) = 3 .
 Step 8: Substitute back.
∑p=3−1+4=6.\sum p = 3 - 1 + 4 = 6 .
 Step 9: The sum is therefore always 66, so it can equal 66 and cannot equal 44, 55 or −1-1. Answer: (1), (2) and (4).
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