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# Help!

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Find a monic cubic polynomial P(x) with integer coefficients such that $$P(\sqrt[3]{2} + 1) = 0$$. (A polynomial is monic if its leading coefficient is 1.)

May 31, 2019

$$P(x) = x^3 + bx^2 + cx + d \\ (2^{1/3}+1)^3 + b(2^{1/3}+1)^2 + c(2^{1/3}+1) +d = 0\\ 3 + 3\left(2^{2/3}+2^{1/3}\right) + b(2^{2/3})+2b(2^{1/3})+b+c(2^{1/3})+c+d=0\\ (3+b)(2^{2/3}) + (3+2b+c)(2^{1/3}) + (3+b+c+d) = 0\\ 3 + b = 0\\ b = -3\\ 3 + 2b + c = 0\\ 3 - 6 + c = 0\\ c = 3\\ 3 + b + c + d = 0\\ 3 - 3 + 3 + d = 0\\ d = -3\\ \therefore \text{The required monic polynomial is } x^3 -3x^2+3x-3=0$$