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HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

Find a nonzero monic polynomial P(x)P(x) with integer coefficients and minimal degree such that P(1−23+43)=0P(1-\sqrt[3]{2}+\sqrt[3]{4})=0. (A polynomial is called monic if its leading coefficient is 1.)

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that (1−23+43)(1+23)=3(1-\sqrt[3]{2}+\sqrt[3]{4})(1+\sqrt[3]{2})=3, so 1−23+43=31+231-\sqrt[3]{2}+\sqrt[3]{4}=\frac{3}{1+\sqrt[3]{2}}. Now, if f(x)=x3−2f(x)=x^{3}-2, we have f(23)=0f(\sqrt[3]{2})=0, so if we let g(x)=f(x−1)=(x−1)3−2=x3−3x2+3x−3g(x)=f(x-1)=(x-1)^{3}-2=x^{3}-3x^{2}+3x-3, then g(1+23)=f(23)=0g(1+\sqrt[3]{2})=f(\sqrt[3]{2})=0. Finally, we let h(x)=g(3x)=27x3−27x2+9x−3h(x)=g\left(\frac{3}{x}\right)=\frac{27}{x^{3}}-\frac{27}{x^{2}}+\frac{9}{x}-3 so h(31+23)=g(1+23)=0h\left(\frac{3}{1+\sqrt[3]{2}}\right)=g(1+\sqrt[3]{2})=0. To make this a monic polynomial, we multiply h(x)h(x) by −x33-\frac{x^{3}}{3} to get x3−3x2+9x−9x^{3}-3x^{2}+9x-9.

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