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Algebra Difficulty 4.9 AIME Find the answer

Let a,b,ca, b, c be the three roots of p(x)=x3+x2−333x−1001p(x)=x^{3}+x^{2}-333 x-1001. Find a3+b3+c3a^{3}+b^{3}+c^{3}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We know that x3+x2−333x−1001=(x−a)(x−b)(x−c)=x3−(a+b+c)x2+(ab+bc+ca)x−abcx^{3}+x^{2}-333 x-1001=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c. Also, (a+b+c)3−3(a+b+c)(ab+bc+ca)+3abc=a3+b3+c3(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)+3 a b c=a^{3}+b^{3}+c^{3}. Thus, a3+b3+c3=(−1)3−3(−1)(−333)+3⋅1001=2003a^{3}+b^{3}+c^{3}=(-1)^{3}-3(-1)(-333)+3 \cdot 1001=2003.

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