Maths Olympiad Prep

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Algebra Difficulty 4.4 AIME Prove it United States

Problem:

Let aa, bb, cc be the three roots of p(x)=x3+x2333x1001p(x) = x^{3} + x^{2} - 333 x - 1001. Find a3+b3+c3a^{3} + b^{3} + c^{3}.

Solution

Solution:

We know that x3+x2333x1001=(xa)(xb)(xc)=x3(a+b+c)x2+(ab+bc+ca)xabcx^{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)33(a+b+c)(ab+bc+ca)+3abc=a3+b3+c3(a + b + c)^{3} - 3(a + b + c)(ab + bc + ca) + 3abc = a^{3} + b^{3} + c^{3}.

Thus,
a3+b3+c3=(1)33(1)(333)+31001=2003. a^{3} + b^{3} + c^{3} = (-1)^{3} - 3(-1)(-333) + 3 \cdot 1001 = 2003.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.