Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore, 
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any ideas?
Thanks a bunch in advance.
 A: Hint:
Notice
\begin{align}
7+5\sqrt{2}&=2\sqrt{2}+3(2)(1)+3(\sqrt{2})(1)+1\\
&=(\sqrt{2})^3+3(\sqrt{2})^2(1)+3(\sqrt{2})(1)^2+(1)^3
\end{align}
Can you recognize this pattern?
A: Suppose $7+5\sqrt{2}=(x+y\sqrt{2})^3 = (x^3+6 x y^2)+(3 x^2 y +2 y^3)\sqrt2 $.
If we're looking for integer solultions, then we must have $7=x^3+6 x y^2$ and $5=3 x^2 y +2 y^3$, because $\sqrt2$ is irrational.
Consider $5=3 x^2 y +2 y^3=(3x^2+2y^2)y$. Since $5$ is prime, $y$ must be $1$ or $5$ because $3x^2+2y^2 \ge 0$. Now $y$ cannot be $5$ because $3x^2+2y^2$ could not be $1$. Thus, $y=1$ and $3x^2+2y^2=5$, which gives $x=\pm 1$.
It is easy to verify that $7+5\sqrt{2}=(1+\sqrt{2})^3$.
A: Remembering that $\sqrt{2}^2=2$ we get, from where you left off.
$$
7+5\sqrt{2} = x^3+3x^2y(\sqrt{2})+3xy^2(2)+y^3(2)(\sqrt{2})\\
            = x^3+6xy^2+3x^2y(\sqrt{2}) +2y^3 (\sqrt{2})\\
            = x^3+6xy^2+(3x^2y+2y^3)(\sqrt{2})
$$
So, we see that $7=x^3+6xy^2$ and $5=3x^2y+2y^3$. Which still looks like a 3 pipe problem untill we stare at the equations for a bit and see that $x=y=1$ is a solution.So $\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$ Which we can verify by calculating $(1+\sqrt{2})^3$.
