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Problem 5

Clearly $ \mathbb{Q}(\alpha)\subset \mathbb{Q}(2^{\frac{1}{3}}) $. But also,

$ 2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}} $

So $ \mathbb{Q}(2^{\frac{1}{3}}) \subset \mathbb{Q}(\alpha) $ and $ \mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha) $.

Thus $ 3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}| $.


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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

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