The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
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$\begingroup$Hint use $a^3+b^3=(a+b)^3-3ab(a+b)$
$\endgroup$ 2 $\begingroup$- $7-5\sqrt 2 = 1-3\sqrt{2} + 6 - 2 \sqrt{2} = 1^3+\dbinom{3}{1}(-\sqrt{2})+\dbinom{3}{2}*(-\sqrt{2})^2+(-\sqrt{2})^3=(1-\sqrt{2})^3$ (By binomial theorem:)
- $7+5\sqrt 2 = 1+3\sqrt{2} + 6 + 2 \sqrt{2} = (1+\sqrt{2})^3$
If the sum is $2$, then $7 \pm 5\sqrt 2$ must be of the form $1 \pm b\sqrt 2$ for some rational $b$ (here it must be an integer because the ring of integers of $\Bbb Q(\sqrt 2)$ is $\Bbb Z[\sqrt 2]$).
Indeed after developing the cube you get $(7 \pm 5 \sqrt 2) = (1+6b^2)+(3b+2b^3)\sqrt 2$. From $1+6b^2 = 7$ and $3b+2b^3 = 5$ you get a solution $b=1$, so $7 \pm 5 \sqrt 2 = (1 \pm \sqrt 2)^3$ and then the simplification is straightforward.
If you don't know that the sum is going to be $2$ but you still think it's going to be a rational number, then it's going to be an integer (because algebraic integers form a ring), so you can make some basic approximations of the sum and get only a few integer values to test.
$\endgroup$ $\begingroup$More generally, if $a + b \sqrt{2}$ (with $a$, $b$ integers) has a cube root in $\mathbb Z[\sqrt{2}]$, the norm $N(a + b \sqrt{2}) = a^2 - 2 b^2 \sqrt{2}$ must be the cube of an integer, say $m^3$, and then that cube root would be $x + y \sqrt{2}$ with $x^2 - 2 y^2 = m$. From $(x+y \sqrt{2})^3 = a + b \sqrt{2}$ we get $x^3 + 6 x y^2 = 4 x^3 - 3 m x = a$. In particular, $x$ must be a divisor of $a$, which brings you down to a finite (and hopefully not too big) list of possiblities that are easily checked.
$\endgroup$ $\begingroup$Frequently, numbers of this kind are obtained from Cardano's formula for the roots of a cubic equation $x^3+px+q=0$, that is, $$ \sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}} $$ so you have $$ -\frac{q}{2}=7, \qquad \frac{q^2}{4}+\frac{p^3}{27}=50, $$ that easily gives $$ q=-14,\qquad p=3. $$ Therefore the equation is $$ x^3+3x-14=0 $$ which has a single real root, precisely $2$.
$\endgroup$ $\begingroup$By direct calculation we have the identity$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac).$
Then set $a=\sqrt[3]{7+5\sqrt{2}},b=\sqrt[3]{7-5\sqrt{2}}$ and $c=-2$. Note that $ab=-1$ and $a^3+b^3=14$, so $a^3+b^3+c^3=3abc$.
Since $(a-b)^2+(b-c)^2+(b-c)^2>0\Longleftrightarrow a^2+b^2+c^2-ab-bc-ac>0$ holds for any three distinct real numbers, we have that $a+b+c=0$, which means $$\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}=2.$$
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