Question 216·Hard·Advanced Math

Use $x=0$ because the $x$ terms disappear. Check both original denominators:

Neither is $0$, so the original expression is defined there.

Type the given product as $f(x)$ in Desmos. Then type $x_1=0$, $f(x_1)$, and the choices with $x$ changed to $x_1$. Desmos gives $1$ for the original; the three nonconstant choices give $-2$, $2$, and $4$. Each fails at an allowed input, so only the constant $1$ can be equivalent. The equivalent expression is $1$. Choice D.

The first denominator is a perfect square, and the second is always positive:

A square cannot be negative, so only $x=2$ is excluded. Keep that restriction even if factors cancel.

$x^3-8$ is a difference of cubes because $8=2^3$. Use $a^3-b^3=(a-b)(a^2+ab+b^2)$, keeping the plus sign in the middle factor:

Replace the numerator and first denominator with their factored forms. Now the matching factors are visible:

Multiply across the two fractions and pair the two copies of $x-2$ on top:

Cancel only whole, nonzero factors; keep the original restriction. Both top and bottom are identical, and $x\ne2$ keeps them nonzero:

So the expression equals $1$ everywhere it is defined. Choice D.