Answer
$2c^2-4c+8$
Work Step by Step
Using $A=bh$ or the formula for the area, $A$, of a rectangle, where $b$ is the base and $h$ is the height, then with $A=2c^3+16$ and $b=c+2$,
$$\begin{aligned}
2c^3+16&=(c+2)h
\\
h&=\frac{2c^3+16}{c+2}
\\
h&=\frac{2(c^3+8)}{c+2}
\\
h&=\frac{2(c^3+2^3)}{c+2}
.\end{aligned}$$
Using $x^3+y^3=(x+y)(x^2-xy+y^2)$ or the factoring of the sum of two cubes, the equation above is equivalent to
$$\begin{aligned}
h&=\frac{2(c+2)(c^2-2c+4)}{c+2}
\\&=
2(c^2-2c+4)
&(\text{cancel }c+2)\\&=
2c^2-4c+8
.\end{aligned}$$Hence, the expression for the missing dimension, $h$, is $2c^2-4c+8$.
