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