Expressing $\sqrt[3]{7+5\sqrt{2}}$ in the form $x+y\sqrt{2}$ Express $\sqrt[3]{(7+5\sqrt{2})}$ in the form $x+y\sqrt{2}$ with $x$ and $y$ rational numbers.
I.e. Show that it is $1+\sqrt{2}$.

If it simplifies, then $7+5\sqrt 2$ is a cube $(a+b \sqrt 2)^3$, in the ring of integers of $\Bbb Q(\sqrt 2)$, which is $\Bbb Z[\sqrt 2]$, so $a$ and $b$ must be integers (sometimes you can only deduce that $2a,a+b,2b$ are integers but it's still very good) 
Moreover, you have $2a = (7+5\sqrt 2)^\frac 13 + (7-5\sqrt 2)^\frac 13$
Since $5\sqrt 2$ is between $7$ and $8$,
The first term is between $2$ and $3$ the second term is between $-1$ and $0$, so the sum has to be $2$ if it's going to be an even integer. So we can bet on $a=1$.
Then writing $7+5\sqrt 3 = (1+b\sqrt 2)^3$ you get $7 = 1+6b^2$, thus $b^2=1$, and you also get $5 = 3b+2b^3 =b(3+2b^2) = b(3+2) = 5b$ so $b=1$.
Since it is compatible with $b^2=1$, it shows that $(1+\sqrt 2)^3 = 7+5\sqrt 2$
