which expression does not factor?\n$m^{3}+1$\n$m^{3}-1$\n$m^{2}+1$\n$m^{2}-1$\ndone

Explanation:

Step1: Recall sum - of - cubes formula

The sum - of - cubes formula is $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. For $m^{3}+1$, where $a = m$ and $b = 1$, we have $m^{3}+1=(m + 1)(m^{2}-m + 1)$.

Step2: Recall difference - of - cubes formula

The difference - of - cubes formula is $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$. For $m^{3}-1$, where $a=m$ and $b = 1$, we have $m^{3}-1=(m - 1)(m^{2}+m + 1)$.

Step3: Recall difference - of - squares formula

The difference - of - squares formula is $a^{2}-b^{2}=(a + b)(a - b)$. For $m^{2}-1$, where $a=m$ and $b = 1$, we have $m^{2}-1=(m + 1)(m - 1)$.

Step4: Analyze $m^{2}+1$

Over the real - number system, there are no two real numbers $a$ and $b$ such that $(a + b)(a - b)=m^{2}+1$. The quadratic formula for $m^{2}+1 = 0$ gives $m=\pm\sqrt{- 1}=\pm i$, which are complex numbers. So $m^{2}+1$ does not factor over the real - number system.

Answer:

$m^{2}+1$