NBHM 2020 PART A Question 1 Solution

Find rational numbers a,b,c such that $(1+\sqrt[3]2)^{-1} = a + b\sqrt[3]2 + c \sqrt[3]2^2$.
Solution: Consider the field extension $\Bbb F = \Bbb{Q}(\sqrt[3]2))$, then $\Bbb F$ is a degree 3 extension of $Q$ with the associated irreducible polynomial $x^3-2$. The set $\{1, \sqrt[3]2), \sqrt[3]2)^2\}$ form a $\Bbb Q$-basis of $\Bbb F$. Hence, it is possible to write $(1+\sqrt[3]2)^{-1} = a + b\sqrt[3]2 + c \sqrt[3]2^2$. Now we have $(1+\sqrt[3]2)(1+\sqrt[3]2)^{-1} = 1$ in $\Bbb F$ and hence $$(1+\sqrt[3]2)(a + b\sqrt[3]2 + c \sqrt[3]2^2) = 1.$$
Expand the left-hand side, we get $$a + b\sqrt[3]2 + c \sqrt[3]2^2 + a\sqrt[3]2 + b\sqrt[3]2^2 + 2c = 1.$$
Now, by equating the coefficients of left and right-hand side we get,
$$a+2c=1, a+b=0, b+c=0$$
In the last step, we have used the fact that the set $\{1, \sqrt[3]2), \sqrt[3]2)^2\}$ form a $\Bbb Q$-basis of $\Bbb F$.
Solving this equations, we get $$a = \frac{1}{3}, b=-\frac{1}{3}, c=\frac{1}{3}.$$
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