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Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.4 AIME Prove it United States

Problem:

Let $a$ , $b$ , $c$ be the three roots of $p(x) = x^{3} + x^{2} - 333 x - 1001$ . Find $a^{3} + b^{3} + c^{3}$ .

Solution

Solution:

We know that $x^{3} + x^{2} - 333 x - 1001 = (x - a)(x - b)(x - c) = x^{3} - (a + b + c) x^{2} + (ab + bc + ca) x - abc$ .

Also, $(a + b + c)^{3} - 3(a + b + c)(ab + bc + ca) + 3abc = a^{3} + b^{3} + c^{3}$ .

Thus,
$a^{3} + b^{3} + c^{3} = (-1)^{3} - 3(-1)(-333) + 3 \cdot 1001 = 2003.$

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