In lecture, we were told that to find $\sqrt[3]{a}$, we use Newton's method as follows: $$ \begin{align} f(x) &= x^3 - a\\ f'(x) &= 3x^2\\ x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)}\\ &= x_n - \frac{{x_n}^3-a}{3{x_n}^2}\\ &=\frac{2}{3}x_n + \frac{a}{3{x_n}^2} \end{align} $$
And for approximating $\frac{1}{a}$,
$$ \begin{align} f(x) &= a - \frac{1}{x}\\ f'(x) &= \frac{1}{x^2}\\ x_{n+1} &= x_n - \frac{x_n}{x_n}\\ &= x_n - (a - \frac{1}{x_n})x^2\\ &=2x_n - a{x_n}^2 \end{align} $$ $$ \\ \\ $$ However, I didn't follow why $f(x) =x^3 - a$ instead of $f(x)=\sqrt[3]{x}$ for the first case and why $f(x)=a - \frac{1}{x}$ instead of $f(x)= \frac{1}{x}$ in the second case.
In each case, the teacher mentioned that the value we're trying to solve for can be realized as the root of $f(x).$ I'm not sure I know what that means in relation to finding a proper f(x).
From what I understand, we are trying to find successively closer $x_{n+1}$ values to the root (where the function is zero) by finding the x intercept of the tangent line of our function at $x_n$ and then repeating that process.
A root for $f(x)$ is simply a value $x_o$ such that $f(x_o)=0$. For the functions you listed each root is the value you seek.
For $f(x)=a-x^3$ if $f(x_o)=a-x_o^3=0$ then $x_o^3=a$ hence $x_o = \sqrt[3]{a}$.
For $f(x)=a-1/x$ if $f(x_o)=a-1/x_o=0$ then $1/x_o=a$ hence $x_o = 1/a$.
The reason not to do what you say is that these work.