Question 198·Hard·Advanced Math

The original denominators are zero at $x=1$ and $x=-1$, so don't test either value. Type $x_{1}=2$ in Desmos; it shows $2$. Use $x_{1}$ instead of $x$ so the following lines print numbers rather than graph curves.

Type the original expression, replacing every $x$ with $x_{1}$: $\frac{x_{1}^{3}-1}{x_{1}-1}-\frac{x_{1}^{2}-1}{x_{1}+1}$. Desmos prints $6$, so an equivalent choice must also give $6$ at $x=2$.

Type the four choices using $x_{1}$. In their listed order, Desmos prints $2$, $4$, $5$, and $6$. Only $x^{2}+2$ matches the original at this input; the other three cannot be equivalent. The factoring approach below confirms that the match holds for every allowed $x$. Choice D.

Because $1=1^{3}$, the first numerator is a difference of cubes. Use $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ with $a=x$ and $b=1$:

The second numerator is a difference of squares. Use $a^{2}-b^{2}=(a-b)(a+b)$ with $a=x$ and $b=1$:

Substitute those factorizations, then cancel the whole factor $x-1$ in the first fraction. It's nonzero because $x\ne1$:

Cancel whole factors, not separate terms of a sum.

Cancel the whole factor $x+1$ in the second fraction; $x\ne-1$ keeps it nonzero:

The canceled factors don't make either excluded input allowed in the original.

Distribute the minus sign across both terms in $(x-1)$, so subtracting $-1$ adds $1$:

Combine like terms, which have the same variable and power:

So for every allowed $x$, the original expression equals $x^{2}+2$. Choice D.