Answer
$a)$ $(f+g)(x)=\sqrt[3]{x}+x-3$
$b)$ $(f-g)(x)=\sqrt[3]{x}-x+3$
$c)$ $(f\cdot g)(x)=x\sqrt[3]{x}-3\sqrt[3]{x}$
$d)$ $\Big(\dfrac{f}{g}\Big)(x)=\dfrac{\sqrt[3]{x}}{x-3}$, where $x\ne3$
Work Step by Step
$f(x)=\sqrt[3]{x};$ $g(x)=x-3$
For each case, replace $f(x)$ by $\sqrt[3]{x}$ and $g(x)$ by $x-3$ and simplify if possible:
$a)$ $(f+g)(x)$
$f(x)+g(x)=(\sqrt[3]{x})+(x-3)=\sqrt[3]{x}+x-3$
$b)$ $(f-g)(x)$
$f(x)-g(x)=(\sqrt[3]{x})-(x-3)=\sqrt[3]{x}-x+3$
$c)$ $(f\cdot g)(x)$
$f(x)\cdot g(x)=(\sqrt[3]{x})(x-3)=x\sqrt[3]{x}-3\sqrt[3]{x}$
$d)$ $\Big(\dfrac{f}{g}\Big)(x)$
$\dfrac{f(x)}{g(x)}=\dfrac{\sqrt[3]{x}}{x-3}$, where $x\ne3$