Simplifications and domain conditions with roots

In each radicand the largest perfect square is separated: $72=36\cdot 2$, $50=25\cdot 2$ and $8=4\cdot 2$. The root of a product of factors greater than or equal to zero is the product of the roots, therefore the square comes out of the radical sign and each term becomes a multiple of $\sqrt{2}$.

The three terms contain the same factor $\sqrt{2}$, hence they are collected like similar monomials: the coefficients are added with their signs and one obtains $(6+5-2)\sqrt{2}=9\sqrt{2}$.

The first root has an even index, therefore it requires a radicand greater than or equal to zero; since the root is in the denominator, it must also be different from zero, and the two requirements together are equivalent to $x-2>0$. The second root has an even index and is in the numerator, therefore the condition $7-x\ge 0$ is enough.

The two conditions must hold together, hence they are combined in a system. From the first it follows that $x>2$ and from the second it follows that $x\le 7$; the intersection of the two sets is the interval $2<x\le 7$.

The root of a root is the root with the product of the indices, therefore in both cases the indices $3$ and $2$ are multiplied and the overall index is $6$. In the first case the radicand is rewritten as a power: $64=2^6$, hence the sixth root of $2^6$ equals $2$.

In the second case the radicand is the power $a^{12}$ and the base $a$ is positive, therefore the root is rewritten as a power with rational exponent: $\sqrt[6]{a^{12}}=a^{\frac{12}{6}}$. The exponent $\frac{12}{6}$ simplifies to $2$, hence the result is $a^2$.