## Algebra 1

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.) $2^3=2\times2\times2=4\times2=8$ $3^3=3\times3\times3=9\times3=27$