Answer
$(f+g)(x)=\sqrt[3] x+x-3$
$(f−g)(x)=\sqrt[3] x−x+3$
$(f×g)(x)=(\sqrt[3] x)×(x-3)=x\sqrt[3] x-3\sqrt[3] x$
$(\frac{f}{g})(x)=\frac{\sqrt[3] x}{x-3}$
Work Step by Step
If $f(x)=\sqrt[3] x$ and $g(x)=x-3$, then
$(f+g)(x)=f(x)+g(x)=\sqrt[3] x+x-3$
$(f−g)(x)=f(x)−g(x)=\sqrt[3] x−x+3$
$(f×g)(x)=f(x)×g(x)=(\sqrt[3] x)×(x-3)=x\sqrt[3] x-3\sqrt[3] x$
$(\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{\sqrt[3] x}{x-3}$
