The set of numbers of the following form that are cubes

Question

I have found this problem in a 10th grade textbook and it's given me headaches trying to solve it. It says, determine the set:

$$ A = \left \{ x \in \mathbb Z| \root3\of{\frac{7x+2}{x+5}} \in \mathbb Z\right \} $$

So I have to find a condition for x so that the expression under the radical is a perfect cube. I remember solving these kind of exercises with perfect squares, but I can't figure this one out.

Answer 1

Let $ \root3\of{\frac{7x+2}{x+5}} = N$

$\frac {7x+2}{x+5} = N^3$

$\frac {7x + 35 - 33}{x + 5} = N^3$

$\frac {7x + 35}{x+5} - \frac {33}{x+5} = N^3$

$7 - \frac {33}{x+5} = N^3 \in \mathbb Z$.

So $x+5$ divides $33$. So $x+5 =\pm 1, \pm 3, \pm 11$ or $\pm 33$.

That means $N^3 = 7-\frac {33}{x+5} = 8,10,18,40,6,4, -4, -26$.

The only one of those that is a perfect cube is $N^2 = 7-\frac {33}{x+5} = 8$ and $\frac {33}{x+5} = -1$.

So $x + 5 = -33$ and $x = -38$.

And check: $\root3\of{\frac{-7*38+2}{-38+5}}= \sqrt[3]{\frac {264}{-33}}=\sqrt[3] 8 = 2$.

Answer 2

Hint: for "large" $x$, the radical is between $6$ and $8$

Answer 3

The key point is that if $\sqrt[3]{a}$ is an integer, then $a$ is an integer, because $a$ is the cube of some integer, and the cube of an integer is always an integer.

Therefore, we conclude that $\frac{7x+2}{x+5}$ is an integer. Write this as $\frac{(7x+35) - 33}{x+5} = 7 - \frac{33}{x+5}$.

Now, if $b$ is an integer, so is $7-b$. From this, we see that $\frac{33}{x+5}$ is an integer.

Consequently, $x+5$ is a (not necessarily positive) divisor of $33$. And how many such divisors of $33$ are there? You get the corresponding values of $x$ and test if $7-\frac{33}{x+5}$ is a perfect cube or not.