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| − | For convenience, let's use <math>a, b, c</math> instead of <math>\alpha, \beta, \gamma</math>. Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc</math> and <math>k = -abc</math>. Then, our polynomial becomes <math>P(x) = x^3 - (a+b+c)x^2 + jx + k</math>.
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| − | Note that we want to compute <math>-\frac{j}{k}</math>.
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| − | From the given information, we know that the coefficient of the <math>x^2</math> term is <math>6</math>, and we also know that <math>P(-1) = -33</math>, or in other words, <math>-j + k = -26</math>. By Newton's Sums (since we are given <math>a^3 + b^3 + c^3</math>), we also find that <math>6j + k = 43</math>. Solving this system, we find that <math>(j, k) \in (\frac{69}{7}, -\frac{113}{7})</math>. Thus, <math>\frac{j}{-k} = \frac{69}{113}</math>, so our final answer is <math>69 + 113 = \boxed{182}</math>.
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