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Let alpha, beta, gamma, delta be the four roots of x⁴ - x³ - x² - 1 = 0. Define p(x) = x⁶ - x⁵ - x³ - x² - x. Then which of the following values cannot be the value of p(alpha) + p(beta) + p(gamma) + p(delta)?

1. 4
2. 5
3. 6
4. 7

Correct answer: 7

Solution

Using alpha⁴ = alpha³ + alpha² + 1, compute alpha⁵ and alpha⁶ in terms of lower powers, substitute into p(alpha), and simplify to get p(alpha) = alpha² - alpha + 1. Summing over all four roots and using Vieta's formulas gives the total sum = 6, which is always fixed — so 4, 5, and 7 cannot be achieved.

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