Rewritting a messy cubic root Is there a really quick way of showing that:
$$\sqrt[3]{49-25\sqrt{2}}$$
Can be written in the form:
$$a+b\sqrt{2}$$
Is there a way to generalize which integers $a$ and $b$ can be rewritten in such a way?
 A: If we assume $\sqrt[3]{45 - 29 \sqrt{2}} = a + b \sqrt{2}$ where $a$ and $b$ are integers, then cubing both sides and expanding gives us $a^3 + 3 a^2 b \sqrt{2} + 6 a b^2 + 2 b^3 \sqrt{2}$, so we want to solve $$\eqalign{a^3 + 6 a b^2 &= 45\cr
3 a^2 b + 2 b^3 &= -29\cr}$$
Now $b$ divides $3 a^2 b + 2 b^3$ and $29$ is prime, so $b$ must be $\pm 1$.
Moreover $b = +1$ would make $3 a^2 + 2 b^3 > 0$, so we must have $b = -1$.
Then the second equation becomes $-3 a^2 - 2 = -29$ or $a^2 = 9$, so $a = \pm 3$.  Finally the first equation becomes $\pm 27 \pm 18 = 45$ which is true for $a=+3$.  Thus the answer is $3 - \sqrt{2}$.  
EDIT: The problem keeps getting edited, but here are some more general considerations.
For $\sqrt[3]{A + B \sqrt{m}} = a + b \sqrt{m}$ where $m$ is a square-free integer, 
you need
$$ \eqalign{ a (a^2 + 3 m b^2) &= A\cr
             b (3 a^2 + m b^2) &= B\cr}$$ 
so $a$ divides $A$ and $b$ divides $B$.   That reduces you to finitely many
possibilities.  To cut it down further you might note that $a \equiv A \mod 3$ and $mb\equiv B \mod 3$, and (if $m > 0$) $a$ and $b$ have the same signs as $A$ and $B$ respectively.
Also, $$ A^2 - B^2 m = (a^2 - b^2 m)^3$$
so a necessary condition for this to work is that $A^2 - B^2 m$ is the cube of an integer.  For example, $49^2 - 2 \times 25^2 = 1151 $ is not the cube of an integer, so $49 - 25 \sqrt{2}$ won't work.
