Use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ to factor $x^3 - 8$, and recognize $x^2 - 4x + 4$ as $(x-2)^2$.

After you rewrite the whole expression with factors, look for the same factors in any numerator and denominator and cancel them. Then see what is left.

In Desmos, enter the function

y1 = (x^3 - 8)/(x^2 - 4x + 4) * (x - 2)/(x^2 + 2x + 4)

You should see a horizontal line (with possibly a missing point at $x=2$ where the original expression is undefined). Note the constant $y$-value of this line.

Now enter each option as its own function: y2 = x - 2, y3 = x + 2, and y4 = x^2 - 2x + 4. Visually compare their graphs with y1. The correct choice is the one whose graph matches the horizontal line of y1 for all $x$ in the domain of the original expression (except the hole at $x=2$).

First, factor the pieces of the expression:

• $x^3 - 8$ is a difference of cubes. Using $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a=x$ and $b=2$:

• $x^2 - 4x + 4$ is a perfect square trinomial:

• The quadratic $x^2 + 2x + 4$ does not factor nicely over the reals, so leave it as is.

Now rewrite the original expression with these factorizations:

Now look for factors that appear in both a numerator and a denominator:

• There is a factor of $(x-2)$ in the first numerator and $(x-2)^2$ in the first denominator. One $(x-2)$ cancels, leaving a single $(x-2)$ in the denominator.
• There is a factor of $(x^2+2x+4)$ in the first numerator and in the second denominator; these cancel completely.
• The remaining $(x-2)$ in the denominator cancels with the $(x-2)$ in the second numerator.

Writing this out:

for all $x$ where the original expression is defined (note $x \neq 2$ because of the original denominators). So the entire expression is equivalent to $1$.