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Let $\alpha = \sqrt[3]{2}$

write $\frac{\alpha^2 -1}{\alpha + 2}$ as a polynomial in $\alpha$ with rational coefficient.

First I write the following:

$\frac{\alpha^2 -1}{\alpha + 2} = \frac{2^{2/3} - 1}{2^{1/3} + 2}$

Then I proceed to manipulate the R.H.S hoping to achieve all rational terms. But it's been a mess so far. Have interpret the question wrongly or is there some algebra trick here I need to do.

Any help or insight is deeply appreciated.

2 Answers 2

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The problem is the denominator, so we can do

$$\frac{\alpha^2 -1}{\alpha + 2}=\frac{(\alpha^2-1)(\alpha^2-2\alpha+4)}{(\alpha+2)(\alpha^2-2\alpha+4)}$$ To make it a polynomial. Now, using the sum of cubes formula, it simplifies to $$\frac{(\alpha^2-1)(\alpha^2-2\alpha+4)}{10}$$As $\alpha=\sqrt[3]{2}$. Done!

There are probably other ways, too.

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    Wow, nice !. But can I ask whats the motivation/intuition behind multiplying top and bottom by the proposed factor ? – 2017-02-23
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    @some1fromhell We want to make the denominator an integer. – 2017-02-23
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    I see, thank you so much. – 2017-02-23
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We have $$ \alpha^3-2=0 \implies \alpha^2+8=10 \implies \frac{1}{\alpha+2}=\frac{\alpha^2-2\alpha+4}{10} $$ Thus $$ \frac{\alpha^2 -1}{\alpha + 2}=\frac{(\alpha^2-1)(\alpha^2-2\alpha+4)}{10}=\frac {1}{10}\,{\alpha}^{4}-\frac{1}{5}\,{\alpha}^{3}+\frac{3}{10}\,{\alpha}^{2}+\frac{1}{ 5}\,\alpha-\frac{2}{5} $$

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    You're wrong, @Leox. Note that $\alpha^3=2$, not $\alpha^2=2$. – 2017-02-23