#### Answer

C) 16

#### Work Step by Step

Use mental math to determine that choices a, b and d all have cube roots that are integers. Verify.
a) If a = -3
$(-3)^3=-3\times(-3)\times(-3)=9\times(-3)=-27\checkmark$
b) If a = -2
$(-2)^3=-2\times(-2)\times(-2)=4\times(-2)=-8\checkmark$
d) if a = 4
$4^3=4\times4\times4=16\times4=64\checkmark$
Since a, b and d have been eliminated, c is the correct choice. There is no integer that equals 16 when cubed.
As shown below, 16 falls between $2^3$ and $3^3$, so $\sqrt[3] {16}$ is a fractional number between 2 and 3. (Fractional numbers are not integers.)
$2^3=2\times2\times2=4\times2=8$
$3^3=3\times3\times3=9\times3=27$