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I came across this question in a text book while preparing for return to university. It has been a few years since I flexed the old grey matter and I have not been able to arrive at the solution. Any help would be much appreciated. The question is:

Express the following in the form $ x + y \sqrt{2}$ with x and y rational numbers: $\sqrt[3]{(7+5\sqrt{2})}$

B.O.B gives the answer as:

$1+\sqrt{2}$

When this expression is cubed it does equal the original expression but I am unsure how to arrive at the solution.

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    The practical approach is to expect $x,y$ to be naturals. Then we are looking for the cube root of about $14.07$, which is between $2$ and $3$. The only candidate is $1+\sqrt 2$ so try cubing that. Bingo! – 2017-01-06
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    That's equivalent to noticing that $z^6 - 14z^3-1$ has $z^2-2z-1$ as a factor, which is not all that obvious. – 2017-01-06

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Hint: Write $\sqrt[3]{(7+5\sqrt{2})}=x + y \sqrt{2}$, then raise both sides to the third power and solve for $x$ and $y$. You will get:

$x(x^2+6y^2)=7$

$y(3x^2+2y^2)=5$

Then eliminate one of the variables (say $y$) and you will get a polynomial $P(x)=0$. Then you have to look for the rational roots of the polynomial (if such roots exist) with Rational Root Theorem

Of course in this particular case, you might just observe that $x=y=1$ is solution of the system (which is strongly suggested by the fact that both $5$ and $7$ are prime and both $x^2+6y^2$ and $3x^2+2y^2$ are positive).

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    It happens for x=y=1 :P – 2017-01-06
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    It is not easy to get to a polynomial in one variable due to the higher powers here. You either need to call in Cardano on one of them or do something like $6y^2=\frac 7x-x^2$ but that won't give a polynomial directly. Without the observation it is a mess. – 2017-01-06
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    Yes, do that, then square the second equation, plug in (you need just the highest power and the free term), solve, verify the solutions. It is not very simple, but not exactly a mess either. – 2017-01-06
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    This answer does indeed provide an algorithmic solution (+1). But @RossMillikan is also right in that the practicality of this approach hinges on having the simplest of all rationals as a root. For example, eliminating $y$ between the equations gives the bicubic in $x\,$: $$256 x^9 - 1344 x^6 + 2460 x^3 - 1372$$ If $x=1$ happened to *not* be an easy root, there would be a whole lot of rational candidates to try by hand. – 2017-01-06
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    P.S. FWIW it works out easier to eliminate the free terms between the equations (multiply the first by $5$, second by $7$, subtract the two) then be left with a homogeneous equation $5 t^3 - 21 t + 30 t - 14 = 0$ where $t = x/y\,$. – 2017-01-06
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    The bicubic simplifies to $64 x^9 - 336 x^6 + 615 x^3 - 343$, which is somewhat simpler. That's not to say that your solution with the homogenous equation is not way better. I wonder in the general case if $\sqrt[3]{a+b\sqrt 2}$ with $a$ and $b$ integers can ever be written as $x+y\sqrt 2$ for non-integer $x$ and/or $y$. – 2017-01-07
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The problem is that not every expression of the form $\sqrt[3]{a+b\sqrt2}\quad$can be expressed in the form $c+d\sqrt2.\quad\{a,b,c,d\} \in \text{integers}$

I have a feeling that whoever posed the question to you already knew that a solution existed.