<s>[INST] 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.
 [/INST] 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 $(1+\sqrt{2})^3$.
</s>