#### Answer

See the explanation below.

#### Work Step by Step

a.
A root of f is the solution of $f(x)=0.$
This problem is to solve $\quad x^{3}-3x+1=0$,
The solutions are numbers $c$ such that $f(c)=0.$
b.
The problem is solved by equating the RHS's.
$x^{3}=3x+1\ \Leftrightarrow\ x^{3}-3x+1=0$,
which is the same problem as in a. We want $c $ such that $f(c)=0.$
c.
The equation is equivalent to $\quad x^{3}-3x+1=0$,
which is the same problem as in a. We want $c $ such that $f(c)=0.$
d.
The problem is solved by equating the RHS's.
$x^{3}-3x=1\ \Leftrightarrow\ x^{3}-3x+1=0$,
which is the same problem as in a. We want $c $ such that $f(c)=0.$
e.
This is the same problem as in a. We want $c $ such that $f(c)=0.$