If $\displaystyle A$ is the additive inverse of $\displaystyle C$, then 

$\displaystyle C+A = 0$.

It follows by way of the commutative and associative properties that

$\displaystyle A+ (B + C)$

$\displaystyle = (B + C) + A$

$\displaystyle = B + (C + A)$

$\displaystyle = B + 0$

$\displaystyle = B$

If $\displaystyle A$ is the additive inverse of $\displaystyle C$, then 

$\displaystyle C+A = 0$, or, equivalently,

$\displaystyle C = -A$

By way of the distributive property and substitution,

$\displaystyle A (B + C)$

$\displaystyle = A B + A C$

$\displaystyle = A B + A (-A)$

$\displaystyle = A B- A ^{2}$

If $\displaystyle B$ is the multiplicative inverse of $\displaystyle C$, then 

$\displaystyle BC = 1$,

or, equivalently,

$\displaystyle C = \frac{1}{B}$.

By way of substitution and the distributive property,

$\displaystyle A (B + C)$

$\displaystyle = AB + AC$

$\displaystyle = AB + A \cdot \frac{1}{B}$

$\displaystyle = AB + \frac{A}{B}$

If $\displaystyle A$ is the multiplicative inverse of $\displaystyle C$, then 

$\displaystyle CA = 1$. 

By way of the commutative and associative properties, substitution, and the identity property of multiplication:

$\displaystyle A\left (BC \right ) = \left (BC \right ) A = B (CA) = B \cdot 1 = B$

According to the order of operations, any operations within parentheses must be performed first. In expression (II), this is the addition; in expression (III), this is the multiplication.

Expression (I) does not have any parentheses, so, by the order of operations, in the absence of grouping symbols, multiplication precedes addition.

Therefore, the correct response is I and III only.

By the distributive property, 

$\displaystyle A (B + C) = AB + AC$ 

$\displaystyle A$ is the multiplicative inverse of $\displaystyle C$, meaning that, by defintion, $\displaystyle AC = 1$, so 

$\displaystyle AB + AC = AB + 1$.

$\displaystyle AB + 1$ is the correct choice.

First, evaluate $\displaystyle 5 \bigstar 2$. Since $\displaystyle 5 > 2$, use the defintion of $\displaystyle a \bigstar b$ for the case $\displaystyle a > b$:

$\displaystyle a \bigstar b = a-b$

$\displaystyle 5 \bigstar 2 = 5 - 2 = 3$.

Now, evaluate $\displaystyle 2 \bigstar 5$. Since $\displaystyle 2 < 5$, use the defintion of $\displaystyle a \bigstar b$ for the case $\displaystyle a< b \;$:

$\displaystyle a \bigstar b = ab$

$\displaystyle 2 \bigstar 5 = 2 \cdot 5 = 10$

The product of $\displaystyle 5 \bigstar 2$ and $\displaystyle 2 \bigstar 5$ is $\displaystyle 3 \cdot 10 = 30$.

$\displaystyle (-8) \bigstar (-8)$ and $\displaystyle 8 \bigstar 8$ are both calculated by using the defintion of $\displaystyle a \bigstar b$ for the case $\displaystyle a = b \;$:

$\displaystyle a \bigstar b = ab$

$\displaystyle (-8) \bigstar (-8) = -8 \cdot (-8) = 64$

$\displaystyle 8 \bigstar 8 = 8 \cdot 8= 64$

Their quotient is $\displaystyle 64 \div 64 = 1$.

Addition and multiplication are both commutative, which means that a sum or product has the same value regardless of the order in which the addends or factors are written. The diagram

$\displaystyle \bigcirc + \bigtriangleup = \bigtriangleup + \bigcirc$

is the one that demonstrates this for addition.

$\displaystyle (-5) \bigstar (-5)$ can be evaluated using the definition of $\displaystyle a \bigstar b$ for the case of both $\displaystyle a$ and $\displaystyle b$ being negative:

$\displaystyle a \bigstar b = a+ b$

$\displaystyle (-5) \bigstar (-5) = -5 + (-5) = -10$

$\displaystyle 5 \bigstar 5$ can be evaluated using the definition of $\displaystyle a \bigstar b$ for the case of $\displaystyle a$ and $\displaystyle b$ not both being negative:

$\displaystyle a \bigstar b = a- b$

$\displaystyle 5 \bigstar 5 = 5-5 = 0$

The quotient: $\displaystyle -10 \div 0$, which is undefined, as zero cannot be taken as a divisor.