Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $

Question

Simplify $$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$

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Attempt:

$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}}{ (3 + \sqrt[3]{4}) - \sqrt[3]{2}} $$ $$ = \frac{ (\sqrt[3]{16} - 1) [(3 + \sqrt[3]{4}) - \sqrt[3]{2}]}{ (3 + \sqrt[3]{4})^{2} - 2^{2/3}} $$ $$ = \frac{ 3 \sqrt[3]{16} - 3\sqrt[3]{4} + \sqrt[3]{2} + 1}{ (9 + 5 \sqrt[3]{4} + \sqrt[3]{16}) } $$

From here on I don't know how to continue.

I can let $a = \sqrt[3]{2}$, but still cannot do anything.

Answer 1

Let $x=\sqrt[3]2$ then we have $${x^4-1\over x^2+x+3}={x^6-x^2\over x(x^3+x^2+3x)}={4-x^2\over x(2+x^2+3x)}= {(2-x)(2+x)\over x(x+2)(x+1) }$$

$$ = {2-x\over x^2+x}= {(2-x)(x-1)\over x(x+1)(x-1)}= {(2-x)(x-1)\over x^3-x}$$

$$= {(2-x)(x-1)\over 2-x} = x-1$$

Edit: but other solution is much nicer then this one.

Answer 2

With the same notation as in the other answer, i.e. $x = \sqrt[3] 2$, noting that $x^3+1=3,$ you can write your quantity as

\begin{eqnarray} \frac{x^4-1}{x^2+x+3} &=& \frac{(x-1)(x^3+x^2+x+1)}{x^2+x+x^3+1}=\\ &=&x-1. \end{eqnarray}

Answer 3

Let $x=\sqrt[3]{2}$ then we have $\frac{x^4-1}{x^2+x+3}$

Multiply by $x-1$

$\frac{x^4-1}{x^2+x+3}\cdot\frac{x-1}{x-1}=\frac{(x^4-1)(x-1)}{x^3+2x-3}=\frac{(x^4-1)(x-1)}{2+2x-3}=\frac{(x^4-1)(x-1)}{2x-1}$

Multiply by $4x^2+2x+1$

$\frac{(x^4-1)(x-1)(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{8x^3-1}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{15}=\frac{1 + x + 2 x^2 - 4 x^3 - x^4 - x^5 - 2 x^6 + 4 x^7}{15}=\frac{15\sqrt[3]{2}-15}{15}=\sqrt[3]{2}-1$

Answer 4

This is a redo of a previously posted answer. This one attempts to be more approachable to readers without knowledge of finite field extensions.

The $\sqrt[3]{16}$ and $\sqrt[3]{27}$ are distractions. Write these as $2\sqrt[3]{2}$ and $3$. So you have $$\frac{-1+2\sqrt[3]{2}}{2+\sqrt[3]{2}+\sqrt[3]{4}}$$ All five terms are in the form $a\sqrt[3]{2^n}$. (For two of the terms, $n=0$.) It would be reasonable to suspect that the result could be written as a sum of such terms. (In fact, finite field theory guarantees this, but it's something you might guess at even without knowing that.)

So you might presume $$\begin{align} \frac{-1+2\sqrt[3]{2}}{3+\sqrt[3]{2}+\sqrt[3]{4}} &=x+y\sqrt[3]{2}+z\sqrt[3]{4}\\ -1+2\sqrt[3]{2}&=\left(3+\sqrt[3]{2}+\sqrt[3]{4}\right)\left(x+y\sqrt[3]{2}+z\sqrt[3]{4}\right)\\ -1+2\sqrt[3]{2}&=3x+3y\sqrt[3]{2}+3z\sqrt[3]{4}+x\sqrt[3]{2}+y\sqrt[3]{4}+2z+x\sqrt[3]{4}+2y+2z\sqrt[3]{2} \end{align}$$ Now equate the parts without cube roots of $2$ or $4$. $$-1=3x+2y+2z$$ And equate the parts with $\sqrt[3]{2}$. $$2=x+3y+2z$$ And equate the parts with $\sqrt[3]{4}$. $$0=x+y+3z$$

This is a system of three linear equations in three unknowns. One way to solve it is with row reduction:

$$ \begin{bmatrix} 3&2&2&-1\\ 1&3&2&2\\ 1&1&3&0 \end{bmatrix} \to \begin{bmatrix} 1&1&3&0\\ 1&3&2&2\\ 3&2&2&-1 \end{bmatrix} \to \begin{bmatrix} 1&1&3&0\\ 0&2&-1&2\\ 0&-1&-7&-1 \end{bmatrix} \to\cdots $$

$$ \cdots\to \begin{bmatrix} 1&1&3&0\\ 0&1&7&1\\ 0&2&-1&2 \end{bmatrix} \to \begin{bmatrix} 1&1&3&0\\ 0&1&7&1\\ 0&0&-15&0\\ \end{bmatrix} \to \begin{bmatrix} 1&1&0&0\\ 0&1&0&1\\ 0&0&1&0\\ \end{bmatrix} \to \begin{bmatrix} 1&0&0&-1\\ 0&1&0&1\\ 0&0&1&0\\ \end{bmatrix} $$ So the solution is $x=-1$, $y=1$, and $z=0$. That is, we get $-1+\sqrt[3]{2}$.

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Here is the original posted answer.

I think the following is a different approach than the answers thus far.

There is a field extension over $\mathbb{Q}$with vector space basis $\{1, \sqrt[3]{2}, \sqrt[3]{4}\}=:\{u,v,w\}$.

You want to simplify $\frac{2v-u}{3u+v+w}$. What is the inverse of $3u+v+w$? Well, the matrix corresponding to multiplication by $3u+v+w$ is $$\begin{bmatrix}3&2&2\\1&3&2\\1&1&3\end{bmatrix}$$ Invert this matrix and you get $$\frac{1}{15}\begin{bmatrix}7&-4&-2\\-1&7&-4\\-2&-1&7\end{bmatrix}$$

So $\frac{2v-u}{3u+v+w}$ corresponds to $$\frac{1}{15}\begin{bmatrix}7&-4&-2\\-1&7&-4\\-2&-1&7\end{bmatrix}\begin{bmatrix}-1\\2\\0\end{bmatrix}=\begin{bmatrix}-1\\1\\0\end{bmatrix}$$

So you get $-u+v$. That is, you get $-1+\sqrt[3]{2}$.