Let a, b, c be the roots of the cubic polynomial
x^3 - 3x^2 + px + 1 = 0.
Given that a^3 + b^3 + c^3 = 6,
find the value of p.
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From the cubic polynomial we know that:
abc=-1,ab+bc+ca=p,a+b+c=3.
But we also have:
a^3+b^3+c^3-3abc=(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)
By substituting the first three equations, we will have:
6-3(-1)=(3)\left((3)^2-3(p)\right)
9=27-9p
p=2
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