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Find a cubic polynomial with integer coefficients that has  $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

(b) Prove that  $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.

 Sep 15, 2020
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Since are looking for a cubic polynomial, let it be ax^3 + bx^2 + cx + d = 0.  So

 

\(a(\sqrt[3]{2} + \sqrt[3]{4})^3 + b(\sqrt[3]{2} + \sqrt[3]{4})^2 + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

 

This expands to

 

\(a (2 + 3 \sqrt[3]{2^2} \sqrt[3]{4} + 3 \sqrt[3]{2} \sqrt[3]{4^2} + 4) + b(\sqrt[3]{2^2} + 2 \sqrt[3]{2} \sqrt[3]{4} + \sqrt[3]{4^2}) + c(\sqrt[3]{2} + \sqrt[3]{4}) + d = 0\)

 

Expanding everything out, and comparing the coefficients, we get

 

a + 3b + 3c + d = 16,
-6b + 3c + 3d = -24,
c - 3d = 22,
d = -6.

 

The solution to this system is a = 1, b = 3, c = 4, d = -6, so the cubic is x^3 + 3x^2 + 4x - 6.

 Sep 15, 2020

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