Find the sum of the squares of the roots of the cubic equation $x^3+3x^2+3x=3$.
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1
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2
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3
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4
The equation is x^3 + 3x^2 + 3x - 3 = 0. This is (x+1)^3 - 4 = 0. Let y = x+1, then y^3 = 4. The roots are y1, y2, y3 where y = cube_root(4) * omega^k. x = y - 1. Sum of squares of roots = sum(yi - 1)^2 = sum(yi^2 - 2yi + 1) = sum(yi^2) - 2*sum(yi) + 3. For y^3 - 4 = 0, sum(yi) = 0 and sum(yi*yj) = 0. sum(yi^2) = (sum yi)^2 - 2*sum(yi*yj) = 0. Thus, sum(xi^2) = 0 - 0 + 3 = 3.
By Vieta's formulas for the cubic equation x^3 + 3x^2 + 3x - 3 = 0, the sum of the roots alpha, beta, and gamma is -3, and the sum of their pairwise products is 3. We use the algebraic identity for the sum of the squares of the roots, which is given by the square of the sum of the roots minus twice the sum of the pairwise products. Substituting the known values into the identity gives (-3)^2 - 2 multiplied by 3. This simplifies to 9 - 6, resulting in a final answer of 3.