Finding a representation of the inverse as linear combination

Question

I am struggling with the following: Write $(7+2^{\frac{1}{3}})^{-1} \in \mathbb{Q}(2^{\frac{1}{3}})$ as a $\mathbb{Q}$-Linear combination of $\{1,2^{\frac{1}{3}},2^{\frac{2}{3}}\}$. Hint: This means for to find coefficients $\lambda_i \in \mathbb{Q}$, such that $(7+2^{\frac{1}{3}})^{-1} =\lambda_1 +\lambda_2 2^{\frac{1}{3}} + \lambda_3 2^{\frac{2}{3}}$ holds.

Thoughts/Question: Because of the hint, the first thing I thought was to just solve this linear equation $(7+2^{\frac{1}{3}})^{-1} =\lambda_1 +\lambda_2 2^{\frac{1}{3}} + \lambda_3 2^{\frac{2}{3}}$ , but I wanted to know if there is "a nice way" to solve this problem.

Answer 1

I think the hint is prompting you to write down this equation: $$ 1 = (7+2^{\frac{1}{3}})^{-1} \times (7+2^{\frac{1}{3}}) =(\lambda_1 +\lambda_2 2^{\frac{1}{3}} + \lambda_3 2^{\frac{2}{3}}) \times (7+2^{\frac{1}{3}}).$$

If you expand out the right hand side, you'll get an expression that is some coefficient times $1$, plus some coefficients times $2^{\frac{1}{3}}$, plus some coefficient times $2^{\frac{2}{3}}$. These coefficients will depend on $\lambda_1$, $\lambda_2$ and $\lambda_3$.

But you know that the whole expression is equal $1$! Therefore, the coefficient in front of the $1$ term must be equal to $1$, and the coefficients in front of the $2^{\frac{1}{3}}$ and $2^{\frac{2}{3}}$ terms must be equal to $0$. This should allow you to solve for $\lambda_1$, $\lambda_2$ and $\lambda_3$.

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If you want a "nice" / fun way to solve this, here's a suggestion:

• $m(X) = X^3 - 2$ is the minimal polynomial for $2^{\frac{1}{3}}$ over $\mathbb Q$. In particular, (i) $m(X)$ is irreducible in $\mathbb Q[X]$, and (ii) $m(2^{\frac{1}{3}}) = 0$.
• Let $g(X) = 7 + X$. Since $m(X)$ is irreducible, the greatest common divisor of $g(X)$ and $m(X)$ must be $1$.
• Therefore, by turning the handle on the Euclidean algorithm, you should be able to find polynomials $a(X), b(X) \in \mathbb Q[X]$ such that $a(X)g(X) + b(X)m(X) = 1$.
• If you substitute $2^{\frac{1}{3}}$ for $X$, you find that $a(2^{\frac{1}{3}})g(2^{\frac{1}{3}}) + b(2^{\frac{1}{3}})m(2^{\frac{1}{3}}) = 1$. But $m(2^{\frac{1}{3}}) = 0$ and $g(2^{\frac{1}{3}}) = 7 + 2^{\frac{1}{3}}$. So $a(2^{\frac{1}{3}})(7 + 2^{\frac{1}{3}}) = 1$.
• Hence $(7 + 2^{\frac{1}{3}})^{-1} = a(2^{\frac{1}{3}})$.