Answer
a. $3^{1/2}\gt3^{1/3}$
b. $\sqrt 7+\sqrt {18}\gt\sqrt {7+18}$
Work Step by Step
a. For a base $b\gt1$, $\sqrt[n] b$ will get smaller when $n$ increases, thus we have $3^{1/2}\gt3^{1/3}$
b. Taking the square of $\sqrt a+\sqrt b$, we have
$(\sqrt a+\sqrt b)^2=a+b+2\sqrt {ab}\gt(a+b)$, if $ab\gt0$.
Thus $\sqrt a+\sqrt b\gt\sqrt {a+b}$ and $\sqrt 7+\sqrt {18}\gt\sqrt {7+18}$
