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 $1+\sqrt{2}$.
 A: You can assume that the nested radical can be expressed in $a+b\sqrt{2}$ form. More specifically, we have $$\sqrt[m]{A+B\sqrt[n]{C}}=a+b\sqrt[n]{C}\tag{1}$$
With your question, we have $$\sqrt[3]{7+5\sqrt{2}}=a+b\sqrt{2}\tag{2}$$
Cubing both sides, we get $$7+5\sqrt{2}=(a^3+6ab^2)+(3a^2b+2b^3)\sqrt{2}\tag{3}$$
And equating corresponding coefficients, we get the following system of equations: $$\begin{cases}a^3+6ab^2=7\\3a^2b+2b^3=5\tag{4}\end{cases}$$
Cross multiplying, we get a multi-variate polynomial. Namely, $$5a^3-21a^2b+30ab^2-14b^3=0\tag{5}$$
Dividing both sides by $b^3$, we get: $$5\frac {a^3}{b^3}-21\frac {a^2}{b^2}+30\frac {a}{b}-14=0\tag{6}$$
Which is also equal to $5\left(\frac ab\right)^3-21\left(\frac {a}{b}\right)^2+30\left(\frac {a}{b}\right)-14=0$. Substituting $a/b$ with $x$, we get the cubic polynomial$$5x^3-21x^2+30x-14=0\tag{7}$$ with $x=1$ as an integer root.
Since $a/b=x$, we have $$\frac ab=1\implies a=b\tag{8}$$
So from $(3)$, we have $a^3+6a(a)^2=7\implies a^3+6a^3=7\implies 7a^3=7\implies a=b=1$
$$\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$$
For practice, you can try to denest $\sqrt[3]{2+\sqrt{5}}$
A: $$(1+\sqrt{2})^3=(1+2\sqrt{2}+2)(1+\sqrt{2})=(3+2\sqrt{2})(1+\sqrt{2})=3+3\sqrt{2}+2\sqrt{2}+4$$
$$\therefore (1+\sqrt{2})^3=7+5\sqrt{2}$$
$$\Rightarrow 1+\sqrt{2}=\sqrt[3]{7+5\sqrt{2}}$$
A: 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 $(1+\sqrt 2)^3 = 7+5\sqrt 2$
