The numerator is a cubic (degree 3) polynomial. Consider whether it could be written as a cube of a binomial, like $(x-3)^3$ or $(x+3)^3$.

Once you factor the numerator and see a common $(x-3)$ factor, remember that $x \ne 3$ allows you to cancel that factor with the denominator.

In Desmos, type

f(x) = (x^3 - 9x^2 + 27x - 27) / (x - 3)

This will graph the original rational expression (with a hole at $x = 3$).

On new lines, enter each option:

A(x) = (x - 3)^2 B(x) = (x + 3)^2 C(x) = x^2 - 9x + 27 D(x) = x^2 - 6x - 9

Zoom in and out as needed and compare the graph of $f(x)$ with each of A(x), B(x), C(x), and D(x). The correct choice is the one whose graph coincides with $f(x)$ at all $x$-values except $x=3$, where $f(x)$ has a hole but the quadratic is defined.

You are dividing a cubic polynomial by a linear binomial: the numerator is degree 3 and the denominator is $x-3$. A natural strategy is to factor the numerator so that it includes $(x-3)$ as a factor, then simplify.

Compare the numerator $x^3 - 9x^2 + 27x - 27$ to the standard expansion of $(a-b)^3$:

Set $a = x$ and $b = 3$ and expand $(x-3)^3$:

This shows that the entire numerator is exactly $(x-3)^3$.

Now rewrite the original expression using the factorization you found:

Because $x \ne 3$, the factor $x-3$ in the numerator and denominator can be canceled:

So the expression is equivalent to $(x-3)^2$, which corresponds to choice A.