We know that $\displaystyle x^2$ is always positive for all values of $\displaystyle x$. Therefore $\displaystyle (-x^2)$ would be negative for all values of $\displaystyle x$. From this conclusion, we know:

$\displaystyle -x^2< x^2$

So we have:

$\displaystyle x^2+2>-x^2+2$

$\displaystyle (b)>(a)$

$\displaystyle (b)$ is the greater quantity.

A positive number raised to the third power will be positive, while a negative number raised to the third power will remain negative.

If $\displaystyle x>0$, then $\displaystyle x^3>0$ and $\displaystyle -x^3< 0$.

If $\displaystyle x< 0$, then $\displaystyle x^3< 0$ and $\displaystyle -x^3>0$.

Since we do not know if $\displaystyle x$ is positive or negative, we cannot draw a conclusion about which option is greater.

If $\displaystyle x>0$, then $\displaystyle (b)$ is greater.

If $\displaystyle x< 0$, then $\displaystyle (a)$ is greater.

When $\displaystyle x>0$ we can write:

$\displaystyle x^3>0\ \text{and}\ -x^3< 0$

We know that $\displaystyle x^3>-x^3$ and $\displaystyle x>-x$. Based on this, we can compare the two given quantities.

$\displaystyle x^3+x+1>-x^3-x+1$

$\displaystyle (b)>(a)$

$\displaystyle (b)$ is the greater quantity.

We know that $\displaystyle x$ is greater than $\displaystyle 3$. We can easily test a few values for $\displaystyle x$ to determine if the values are increasing or decreasing.

If $\displaystyle {x=3}$:

$\displaystyle (a)=3^2-6(3)+10=9-18+10=1$

If $\displaystyle x=4$:

$\displaystyle (a)=4^2-6(4)+10=16-24+10=2$

If $\displaystyle x=5$:

$\displaystyle (a)=5^2-6(5)+10=25-30+10=5$

The value of $\displaystyle (a)$ is increasing, with the smallest possible value being $\displaystyle 1$. From this, we know that $\displaystyle (a)>0$, so $\displaystyle (a)>(b)$.

Using the distributive property:

$\displaystyle 3t + 5t = (3 + 5)t = 8t$

and

$\displaystyle 80t - 10t = (80-10)t = 70t$

Using the associative property of multiplication:

$\displaystyle 4 \cdot 4t =\left ( 4 \cdot 4 \right ) t= 16t$

We can rewrite $\displaystyle 16t \div 8$ as $\displaystyle 16t \cdot \frac{1}{8}$; using the commutative and associative properties of multiplication:

$\displaystyle 16t \cdot \frac{1}{8} = \left ( 16 \cdot \frac{1}{8} \right ) \cdot t = 2t$

$\displaystyle 6 + 2t$ is the sum of unlike terms and cannot be simplified.

$\displaystyle 3t + 5t$ is the correct choice.

Depending on the value of $\displaystyle t$, it is possible for either expression to be greater or for both to be equal.

Case 1: $\displaystyle t = 1$

$\displaystyle 5t + 8t + 7 = 5 \cdot 1 + 8 \cdot 1 + 7 = 5 + 8 + 7 = 20$

and 

$\displaystyle 5+ 8t + 7t = 5+ 8 \cdot 1 + 7 \cdot 1 = 5 + 8 + 7 = 20$

So the two are equal.

Case 2: $\displaystyle t = 2$

$\displaystyle 5t + 8t + 7 = 5 \cdot 2 + 8 \cdot 2 + 7 = 10 + 16 + 7 = 33$

and 

$\displaystyle 5+ 8t + 7t = 5+ 8 \cdot 2 + 7 \cdot 2 = 5 + 16 + 14 = 35$

So (B) is greater. 

The correct response is that it cannot be determined which is greater.

$\displaystyle 6t+ 5t + 4t + 3t + 2 t + t = (6+5+4+3+2+1)t = 21t$

$\displaystyle 7t+5t+3t+t = (7 + 5 + 3 + 1)t = 16t$

Since $\displaystyle 21 > 16$, and $\displaystyle t$ is positive,

then by the multiplication property of inequality,

$\displaystyle 21 t > 16t$

making (A) greater regardless of the value of $\displaystyle t$.

$\displaystyle 7t + 11 + 4t = 7t + 4t + 11= (7+4)t + 11 = 11t + 11$

$\displaystyle 7 + 11t + 4 = 11t + 7 + 4 = 11t + 11$

Regardless of the value of $\displaystyle t$, the expressions are equal.

The expression is the sum of two unlike terms, and therefore cannot be further simplified. None of these responses is correct.

$\displaystyle 8t + 7 + 5t + 9 = \left ( 8 + 5 \right ) t + 9+ 7 = 13t + 16$

$\displaystyle 10 + 4t + 9t +5 = (4+9)t + 10 + 5 = 13t + 15$

Since $\displaystyle 16 > 15$, $\displaystyle 13t+ 16 > 13t+ 15$, so (A) is greater regardless of the value of $\displaystyle t$.